How to derive the formula of the sum of this finite series: $sum_{n=M}^N a^n$
$begingroup$
I would like to know how to arrive at the following result that
my teacher wrote on the board. They did not explain how it was done. I am also not sure what this series is called. Is it perhaps a power series?
$$N>M :sum_{n=M}^N a^n = frac{a^M-a^1a^n}{1-a},aneq1$$
$$N>M: sum_{n=M}^N a^n = N-M+1,a=1$$
I am quite lost since my teacher only wrote the above formulae without any derivation. Can someone help me understand why they are true? Thank you!
algebra-precalculus summation
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
$endgroup$
add a comment |
$begingroup$
I would like to know how to arrive at the following result that
my teacher wrote on the board. They did not explain how it was done. I am also not sure what this series is called. Is it perhaps a power series?
$$N>M :sum_{n=M}^N a^n = frac{a^M-a^1a^n}{1-a},aneq1$$
$$N>M: sum_{n=M}^N a^n = N-M+1,a=1$$
I am quite lost since my teacher only wrote the above formulae without any derivation. Can someone help me understand why they are true? Thank you!
algebra-precalculus summation
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
$endgroup$
$begingroup$
Looks like an application of geometric series, i.e. $sum_{i=0}^n a^i = frac{1-a^{n+1}}{1-a}$.
$endgroup$
– TrostAft
Dec 6 '18 at 16:01
add a comment |
$begingroup$
I would like to know how to arrive at the following result that
my teacher wrote on the board. They did not explain how it was done. I am also not sure what this series is called. Is it perhaps a power series?
$$N>M :sum_{n=M}^N a^n = frac{a^M-a^1a^n}{1-a},aneq1$$
$$N>M: sum_{n=M}^N a^n = N-M+1,a=1$$
I am quite lost since my teacher only wrote the above formulae without any derivation. Can someone help me understand why they are true? Thank you!
algebra-precalculus summation
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
$endgroup$
I would like to know how to arrive at the following result that
my teacher wrote on the board. They did not explain how it was done. I am also not sure what this series is called. Is it perhaps a power series?
$$N>M :sum_{n=M}^N a^n = frac{a^M-a^1a^n}{1-a},aneq1$$
$$N>M: sum_{n=M}^N a^n = N-M+1,a=1$$
I am quite lost since my teacher only wrote the above formulae without any derivation. Can someone help me understand why they are true? Thank you!
algebra-precalculus summation
algebra-precalculus summation
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
edited Dec 10 '18 at 12:16
Brahadeesh
6,32942363
6,32942363
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
asked Dec 6 '18 at 15:59
KnowledgeKnowledge
113
113
$begingroup$
Looks like an application of geometric series, i.e. $sum_{i=0}^n a^i = frac{1-a^{n+1}}{1-a}$.
$endgroup$
– TrostAft
Dec 6 '18 at 16:01
add a comment |
$begingroup$
Looks like an application of geometric series, i.e. $sum_{i=0}^n a^i = frac{1-a^{n+1}}{1-a}$.
$endgroup$
– TrostAft
Dec 6 '18 at 16:01
$begingroup$
Looks like an application of geometric series, i.e. $sum_{i=0}^n a^i = frac{1-a^{n+1}}{1-a}$.
$endgroup$
– TrostAft
Dec 6 '18 at 16:01
$begingroup$
Looks like an application of geometric series, i.e. $sum_{i=0}^n a^i = frac{1-a^{n+1}}{1-a}$.
$endgroup$
– TrostAft
Dec 6 '18 at 16:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The first one can be derived by the geometric series
$$sum_{n=M}^N a^n =sum_{n=0}^N a^n-sum_{n=0}^{M-1} a^n $$
the second one is simply
$$sum_{n=M}^N 1 $$
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7.
$endgroup$
– Knowledge
Dec 6 '18 at 16:12
$begingroup$
@DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
$endgroup$
– gimusi
Dec 6 '18 at 16:16
$begingroup$
oh my bad thank you !!!
$endgroup$
– Knowledge
Dec 6 '18 at 16:18
$begingroup$
@DvirIhie You are welcome! Bye
$endgroup$
– gimusi
Dec 6 '18 at 16:19
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first one can be derived by the geometric series
$$sum_{n=M}^N a^n =sum_{n=0}^N a^n-sum_{n=0}^{M-1} a^n $$
the second one is simply
$$sum_{n=M}^N 1 $$
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7.
$endgroup$
– Knowledge
Dec 6 '18 at 16:12
$begingroup$
@DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
$endgroup$
– gimusi
Dec 6 '18 at 16:16
$begingroup$
oh my bad thank you !!!
$endgroup$
– Knowledge
Dec 6 '18 at 16:18
$begingroup$
@DvirIhie You are welcome! Bye
$endgroup$
– gimusi
Dec 6 '18 at 16:19
add a comment |
$begingroup$
The first one can be derived by the geometric series
$$sum_{n=M}^N a^n =sum_{n=0}^N a^n-sum_{n=0}^{M-1} a^n $$
the second one is simply
$$sum_{n=M}^N 1 $$
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7.
$endgroup$
– Knowledge
Dec 6 '18 at 16:12
$begingroup$
@DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
$endgroup$
– gimusi
Dec 6 '18 at 16:16
$begingroup$
oh my bad thank you !!!
$endgroup$
– Knowledge
Dec 6 '18 at 16:18
$begingroup$
@DvirIhie You are welcome! Bye
$endgroup$
– gimusi
Dec 6 '18 at 16:19
add a comment |
$begingroup$
The first one can be derived by the geometric series
$$sum_{n=M}^N a^n =sum_{n=0}^N a^n-sum_{n=0}^{M-1} a^n $$
the second one is simply
$$sum_{n=M}^N 1 $$
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
$endgroup$
The first one can be derived by the geometric series
$$sum_{n=M}^N a^n =sum_{n=0}^N a^n-sum_{n=0}^{M-1} a^n $$
the second one is simply
$$sum_{n=M}^N 1 $$
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
answered Dec 6 '18 at 16:04
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7.
$endgroup$
– Knowledge
Dec 6 '18 at 16:12
$begingroup$
@DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
$endgroup$
– gimusi
Dec 6 '18 at 16:16
$begingroup$
oh my bad thank you !!!
$endgroup$
– Knowledge
Dec 6 '18 at 16:18
$begingroup$
@DvirIhie You are welcome! Bye
$endgroup$
– gimusi
Dec 6 '18 at 16:19
add a comment |
$begingroup$
but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7.
$endgroup$
– Knowledge
Dec 6 '18 at 16:12
$begingroup$
@DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
$endgroup$
– gimusi
Dec 6 '18 at 16:16
$begingroup$
oh my bad thank you !!!
$endgroup$
– Knowledge
Dec 6 '18 at 16:18
$begingroup$
@DvirIhie You are welcome! Bye
$endgroup$
– gimusi
Dec 6 '18 at 16:19
$begingroup$
but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7.
$endgroup$
– Knowledge
Dec 6 '18 at 16:12
$begingroup$
but you wrote from 0 to N i dont think its the same, sum of series for exm. from 5 till 8 its not like from 0 to 5 minus series from 0 to 7.
$endgroup$
– Knowledge
Dec 6 '18 at 16:12
$begingroup$
@DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
$endgroup$
– gimusi
Dec 6 '18 at 16:16
$begingroup$
@DvirIhie But form $5$ to $8$ is like $0$ to $8$ minus $0$ to $4$, that's what I wrote.
$endgroup$
– gimusi
Dec 6 '18 at 16:16
$begingroup$
oh my bad thank you !!!
$endgroup$
– Knowledge
Dec 6 '18 at 16:18
$begingroup$
oh my bad thank you !!!
$endgroup$
– Knowledge
Dec 6 '18 at 16:18
$begingroup$
@DvirIhie You are welcome! Bye
$endgroup$
– gimusi
Dec 6 '18 at 16:19
$begingroup$
@DvirIhie You are welcome! Bye
$endgroup$
– gimusi
Dec 6 '18 at 16:19
add a comment |
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$begingroup$
Looks like an application of geometric series, i.e. $sum_{i=0}^n a^i = frac{1-a^{n+1}}{1-a}$.
$endgroup$
– TrostAft
Dec 6 '18 at 16:01