Binomial Theorem with Three Terms











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$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.










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  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42















up vote
4
down vote

favorite
1












$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.










share|cite|improve this question


















  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42













up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.










share|cite|improve this question













$(x^2 + 2 + frac{1}{x} )^7$



Find the coefficient of $x^8$



Ive tried to combine the $x$ terms and then use the general term of the binomial theorem twice but this does seem to be working.



Does anyone have a method of solving this questions and others similar efficiently?



Thanks.







combinatorics binomial-coefficients






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asked Nov 19 at 10:18









ultralight

376




376








  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42














  • 1




    Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
    – Henry
    Nov 19 at 12:16






  • 1




    A "binomial with three terms" should probably be called a trinomial.
    – Torsten Schoeneberg
    Nov 20 at 8:42








1




1




Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
Nov 19 at 12:16




Perhaps slightly easier to find the coefficient of $x^{15}$ in the expansion of $(x^3+2x+1)^7$
– Henry
Nov 19 at 12:16




1




1




A "binomial with three terms" should probably be called a trinomial.
– Torsten Schoeneberg
Nov 20 at 8:42




A "binomial with three terms" should probably be called a trinomial.
– Torsten Schoeneberg
Nov 20 at 8:42










5 Answers
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up vote
13
down vote



accepted










In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



Thus the coefficient of $x^8$ is $8(35)+21 = 301$






share|cite|improve this answer





















  • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
    – mathnoob
    Nov 19 at 10:47










  • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
    – Mohammad Riazi-Kermani
    Nov 19 at 10:51


















up vote
11
down vote













The multinomial theorem can come to the rescue:
$$
(a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
$$



where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
$$
2i-k=8,qquad i+kle 7
$$

Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





  • $i=4$, $k=0$, $j=3$;


  • $i=5$, $k=2$, $j=0$.


Thus the coefficient is
$$
2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
$$






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    Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
    Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



    Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



    If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



    If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



    If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



    so the answer is $301$.






    share|cite|improve this answer






























      up vote
      2
      down vote













      The answer is 301.



      Just trust your plan of the twofold use of the binomial formula:



      First step



      $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



      Second step



      $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



      Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



      $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



      Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






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        To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



        This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



        7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

        7+4+4, 4+7+4, 4+4+7

        6+6+3, 6+3+6, 3+6+6

        6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



        That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






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          5 Answers
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          5 Answers
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          up vote
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          accepted










          In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



          There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



          Thus the coefficient of $x^8$ is $8(35)+21 = 301$






          share|cite|improve this answer





















          • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
            – mathnoob
            Nov 19 at 10:47










          • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
            – Mohammad Riazi-Kermani
            Nov 19 at 10:51















          up vote
          13
          down vote



          accepted










          In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



          There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



          Thus the coefficient of $x^8$ is $8(35)+21 = 301$






          share|cite|improve this answer





















          • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
            – mathnoob
            Nov 19 at 10:47










          • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
            – Mohammad Riazi-Kermani
            Nov 19 at 10:51













          up vote
          13
          down vote



          accepted







          up vote
          13
          down vote



          accepted






          In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



          There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



          Thus the coefficient of $x^8$ is $8(35)+21 = 301$






          share|cite|improve this answer












          In order to get $x^8$ in the product you have to have either $$x^2 x^2 x^2 x^2 times 2^3$$ or$$ x^2 x^2 x^2 x^2 x^2 (1/x)(1/x)$$



          There are $binom 7 4 $ of the first type and $ binom 7 5$ of the second type.



          Thus the coefficient of $x^8$ is $8(35)+21 = 301$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 10:44









          Mohammad Riazi-Kermani

          40.2k41958




          40.2k41958












          • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
            – mathnoob
            Nov 19 at 10:47










          • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
            – Mohammad Riazi-Kermani
            Nov 19 at 10:51


















          • But in the first type, don't you also have to multiply by $(x^{-1})^3$?
            – mathnoob
            Nov 19 at 10:47










          • @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
            – Mohammad Riazi-Kermani
            Nov 19 at 10:51
















          But in the first type, don't you also have to multiply by $(x^{-1})^3$?
          – mathnoob
          Nov 19 at 10:47




          But in the first type, don't you also have to multiply by $(x^{-1})^3$?
          – mathnoob
          Nov 19 at 10:47












          @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
          – Mohammad Riazi-Kermani
          Nov 19 at 10:51




          @mathnoob No, we only have $7$ parenthesis to multiply and our choices are exhausted by four $x^2$ and three $2$
          – Mohammad Riazi-Kermani
          Nov 19 at 10:51










          up vote
          11
          down vote













          The multinomial theorem can come to the rescue:
          $$
          (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
          $$



          where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



          Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
          $$
          2i-k=8,qquad i+kle 7
          $$

          Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





          • $i=4$, $k=0$, $j=3$;


          • $i=5$, $k=2$, $j=0$.


          Thus the coefficient is
          $$
          2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
          8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
          $$






          share|cite|improve this answer



























            up vote
            11
            down vote













            The multinomial theorem can come to the rescue:
            $$
            (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
            $$



            where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



            Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
            $$
            2i-k=8,qquad i+kle 7
            $$

            Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





            • $i=4$, $k=0$, $j=3$;


            • $i=5$, $k=2$, $j=0$.


            Thus the coefficient is
            $$
            2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
            8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
            $$






            share|cite|improve this answer

























              up vote
              11
              down vote










              up vote
              11
              down vote









              The multinomial theorem can come to the rescue:
              $$
              (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
              $$



              where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



              Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
              $$
              2i-k=8,qquad i+kle 7
              $$

              Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





              • $i=4$, $k=0$, $j=3$;


              • $i=5$, $k=2$, $j=0$.


              Thus the coefficient is
              $$
              2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
              8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
              $$






              share|cite|improve this answer














              The multinomial theorem can come to the rescue:
              $$
              (a+b+c)^n=sum_{i+j+k=n}binom{n}{i,j,k}a^ib^jc^k
              $$



              where $dbinom{n}{i,j,k} = dfrac{n!}{i! , j! , k!}$.



              Here $n=7$, $a=x^2$, $b=2$, $c=x^{-1}$. How can we get $a^ic^k=x^8$? We need
              $$
              2i-k=8,qquad i+kle 7
              $$

              Hence $k=2i-8$ and $3i-8le 7$, so $ige4$ and $ile 5$. Hence we have the cases





              • $i=4$, $k=0$, $j=3$;


              • $i=5$, $k=2$, $j=0$.


              Thus the coefficient is
              $$
              2^3binom{7}{4,3,0}+binom{7}{5,0,2}=
              8frac{7!}{4!,3!,0!}+frac{7!}{5!,0!,2!}=8cdot35+21=301
              $$







              share|cite|improve this answer














              share|cite|improve this answer



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              edited Nov 19 at 21:13

























              answered Nov 19 at 15:26









              egreg

              174k1383198




              174k1383198






















                  up vote
                  5
                  down vote













                  Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                  Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                  Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                  If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                  If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                  If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                  so the answer is $301$.






                  share|cite|improve this answer



























                    up vote
                    5
                    down vote













                    Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                    Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                    Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                    If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                    If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                    If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                    so the answer is $301$.






                    share|cite|improve this answer

























                      up vote
                      5
                      down vote










                      up vote
                      5
                      down vote









                      Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                      Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                      Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                      If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                      If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                      If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                      so the answer is $301$.






                      share|cite|improve this answer














                      Let $$R(x)= left(x^2+2+{1over x}right)^7$$ then we need to find a coeficient at $x^{15}$ for$$x^7R(x)= (x^3+2x+1)^7 $$ $$= sum _{k=0}^7 {7choose k}x^{21-3k}(2x+1)^k$$
                      Clearly if $21-3kgeq 16$ there is no term with $x^{15}$ so $21-3kleq 15$ so $kgeq 2$.



                      Also if $21-3kleq 7$ we have no term with $x^{15}$ so $21-3kgeq 8$ so $3kleq 13$ so $kleq 4$.



                      If $k=2$ we have $${7choose 2}x^{15}(2x+1)^2$$ so the term is $21$



                      If $k=3$ we have $${7choose 3}x^{12}(2x+1)^3$$ so the term is $35cdot 8= 280$



                      If $k=4$ we have $${7choose 4}x^{9}(2x+1)^4$$ there is no trem with $x^{15}$



                      so the answer is $301$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 19 at 15:37









                      Rad80

                      30718




                      30718










                      answered Nov 19 at 10:42









                      greedoid

                      34.4k114488




                      34.4k114488






















                          up vote
                          2
                          down vote













                          The answer is 301.



                          Just trust your plan of the twofold use of the binomial formula:



                          First step



                          $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                          Second step



                          $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                          Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                          $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                          Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






                          share|cite|improve this answer



























                            up vote
                            2
                            down vote













                            The answer is 301.



                            Just trust your plan of the twofold use of the binomial formula:



                            First step



                            $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                            Second step



                            $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                            Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                            $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                            Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






                            share|cite|improve this answer

























                              up vote
                              2
                              down vote










                              up vote
                              2
                              down vote









                              The answer is 301.



                              Just trust your plan of the twofold use of the binomial formula:



                              First step



                              $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                              Second step



                              $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                              Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                              $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                              Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.






                              share|cite|improve this answer














                              The answer is 301.



                              Just trust your plan of the twofold use of the binomial formula:



                              First step



                              $$left((x^2+2)+frac{1}{x}right)^7=sum _{k=0}^7 binom{7}{k} left(x^2+2right)^k x^{k-7}$$



                              Second step



                              $$left(x^2+2right)^k=sum _{m=0}^k 2^{k-m} x^{2 m} binom{k}{m}$$



                              Hence you get a double sum in which the power of $x$ is $2m+k-7$, setting this equal to $8$ we get $k = 15-2m$. This leaves this single sum over $m$



                              $$sum _{m=0}^7 2^{15-3 m} binom{7}{15-2 m} binom{15-2 m}{m}$$



                              Since, for $n, m = 0,1,2,...$ the binomial coefficient $binom{n}{m}$ is zero unless $nge m$ we find $7ge 15-2m to m ge 4$ and $15-2mge m to mle 5$. Hence only the terms with $m=4$ and $m=5$ contribute to the sum giving $280$ and $31$, respectively, the sum of which is $301$.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 19 at 11:29

























                              answered Nov 19 at 11:09









                              Dr. Wolfgang Hintze

                              3,025516




                              3,025516






















                                  up vote
                                  0
                                  down vote













                                  To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                  This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                  7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                  7+4+4, 4+7+4, 4+4+7

                                  6+6+3, 6+3+6, 3+6+6

                                  6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                  That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






                                  share|cite|improve this answer

























                                    up vote
                                    0
                                    down vote













                                    To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                    This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                    7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                    7+4+4, 4+7+4, 4+4+7

                                    6+6+3, 6+3+6, 3+6+6

                                    6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                    That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






                                    share|cite|improve this answer























                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                      This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                      7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                      7+4+4, 4+7+4, 4+4+7

                                      6+6+3, 6+3+6, 3+6+6

                                      6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                      That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).






                                      share|cite|improve this answer












                                      To expand on Henry's comment, this is equivalent to finding the coefficient of $x^{15}$ in $(x^3+x+x+1)^7$. And that is equivalent to finding the number of ways of choosing (with replacement) seven numbers from [3,1,1,0] that add up to 15 (note for the purposes of this counting, the two 1's are distinguishable). In other words, how many length seven sequence of 3's, 1's, 1's, and 0's are there with a sum of 15? Start with the largest number first: if you have zero 3's, then the most you can get is by taking seven 1's, giving you 7, which is too small. With one 3, you can get at most 9. With two 3's, 11. Three 3's, 13. It's not until you get to four 3's that you can get 15, with four 3's and three 1's. There are ${7 choose 4}$ different orderings of the 3's. Since the 1's are distinguishable, and there are two options of which to take each time, that contributes a factor of $2^3=8$. If we have five 3's, that means that the rest have to be 0, so that gives ${7choose 5}$ possibilities. Once you get past five 3's, you're at more than 15 for the total, so that's it: $2^3{7 choose 4}+{7choose 5}$.



                                      This approach can be used more generally. For instance, suppose you want the coefficient of $x^{15}$ for $(x^7+x^6+x^5+x^4+x^3)^3$. You then need to find the number of ways to take from [7,6,5,4,3] with replacement three times and get a sum of 15. You have



                                      7+5+3, 7+3+4, 5+7+3, 5+3+7, 3+7+5, 3+5+7

                                      7+4+4, 4+7+4, 4+4+7

                                      6+6+3, 6+3+6, 3+6+6

                                      6+5+4, 6+4+5, 5+6+4, 5+4+6, 4+6+5, 4+5+6



                                      That's a total of 18, so the coefficient of $x^{15}$ will be 18 (note that each line is just permutations of the same numbers).







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 19 at 18:37









                                      Acccumulation

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