How do you find the sum of ${(- 2)}^{i - 1}$ $(-2)^(i-1)$ from i=1 to 12?

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Answer 1 · 2017-02-19T19:18:19

The sum will equal -1365. How you get this is shown below...

This is a geometric series with a common ratio of -2 and a first term equal to 1 since ${(- 2)}^{1 - 1} = {(- 2)}^{0} = 1$ $(-2)^(1-1) = (-2)^0 = 1$

So, the sum of $n$ $n$ terms is given by

$S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}$ $S_n = (a_1(1-r^n))/(1-r)$

Here, $n$ $n$ = 12 if we want the sum of the first 12 terms, so

$S_{12} = \frac{1 (1 - {(- 2)}^{12})}{1 - (- 2)}$ $S_12 = (1(1-(-2)^12))/(1-(-2))$

$= \frac{1 - 4096}{3} = - \frac{4095}{3} = - 1365$ $=(1-4096)/3 = -4095/3 = -1365$

How do you find the sum of (−2)i−1(-2)^(i-1) from i=1 to 12?