# What is sum_(n=1)^(4) [100(-4)^(n-1)]?

May 22, 2018

The answer is $= - 5100$

#### Explanation:

This is a geometric progression

$r = - 4$

${a}_{n} = 100 \left({\left(- 4\right)}^{n - 1}\right)$

${a}_{n + 1} = 100 \left({\left(- 4\right)}^{n + 1 - 1}\right)$

${a}_{1} = 100$

The sum is

$S = {a}_{1} \frac{1 - {r}^{n + 1}}{1 - r}$

$= 100 \frac{1 - {\left(- 4\right)}^{4}}{1 - \left(- 4\right)}$

$= \frac{100}{5} \left(1 - 256\right)$

$= - 5100$

Jul 1, 2018

Here is an alternative, simple approach. Since there are only 4 terms in the sum, just expand it.

sum_(n=1)^(4) [100(-4)^(n-1)] = ?

There is no $n$ dependence in $100$, so factor it out:

$= 100 {\sum}_{n = 1}^{4} {\left(- 4\right)}^{n - 1}$

Now expand:

$= 100 \left[{\left(- 4\right)}^{1 - 1} + {\left(- 4\right)}^{2 - 1} + {\left(- 4\right)}^{3 - 1} + {\left(- 4\right)}^{4 - 1}\right]$

$= 100 \left[{\left(- 4\right)}^{0} + {\left(- 4\right)}^{1} + {\left(- 4\right)}^{2} + {\left(- 4\right)}^{3}\right]$

$= 100 \left(1 - 4 + 16 - 64\right)$

$= 100 \left(- 51\right)$

$= \textcolor{b l u e}{- 5100}$