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4−12+36−108+324−...+236,196

What is the common trick to compute numbers with the same path ?

Here the path is have x-3x+3(3x)-.....

Thanks!

So the number is

$4 - 3*4 + 3^2*4 - ...... + 3^10*4 = $

$4(1 - 3 + 3^2 - ...... + 3^{10})=$

$4(1 + (-3)^1 +(-3)^2 + ..... + (-3)^{10}) = $

$4frac {1+ (-3)^{11}}{1 - (-3)}=$ (do you know that trick? [1])

$4frac {1 - 3^{11}}{4} = 1 - 177147= -177146$

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[1]$(1 + a + a^2 + ..... + a^n)(1-a)=$

$(1 + a + a^2 + ..... + a^n)-(a + a^2 + a^2 + ..... + a^{n+1})=$

$(1- a^{n+1})$ and therefore:

$ 1 + a + a^2 + ..... + a^n= frac {1 - a^{n+1}}{1-a} $

And $r + ra + ra^2 + ..... + ra^n = rfrac {1 - a^{n+1}}{1-a}$

It's a very handy trick. It called a geometric series

This is a Geometric Series, i.e. a sum of a sequence of numbers where each number is obtained from the previous by multiplying by the same number each time. For example, the sum $1-2+4-8+cdots$ is a Geometric Series since you get each number by multiplication by $-2$ whereas $1+2+4+8+24+72+cdots$ is not Geometric since there is not a consistent number to get to the next number each time.

If you are summing a finite number of geometric terms, the sum is
$$
a+ar+ar^2+cdots= a left( dfrac{1-r^n}{1-r} right),
$$
where $a$ is the first number, $r$ is the number you multiply by, i.e. the common ratio, and $n$ is the number of terms. For example,
$$
1-3+9-27+81= 1 cdot left(dfrac{1-(-3)^5}{1-(-3)} right)=
$$

If you use an infinite amount of terms, then the sum is
$$
a+ar+ar^2+cdots+ar^3+cdots= dfrac{a}{1-r},
$$
assuming that the number $r$ you multiply by has the property that $-1<r<1$, otherwise the sum will not converge. For example,
$$
3- dfrac{3}{2} + dfrac{3}{4} - dfrac{3}{8} + cdots = dfrac{3}{1-(-1/2)}= dfrac{3}{3/2}= 3.
$$
For more on this, check out the Wiki page: Geometric Series