Question

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

Answer 1

The answer is `=-5100`

Explanation:

This is a geometric progression

`r=-4`

`a_n=100((-4)^(n-1))`

`a_(n+1)=100((-4)^(n+1-1))`

`a_1=100`

The sum is

`S=a_1(1-r^(n+1))/(1-r)`

`=100(1-(-4)^4)/(1-(-4))`

`=100/5(1-256)`

`=-5100`

Answer 2

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

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

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

`= 100 sum_(n=1)^(4) (-4)^(n-1)`

Now expand:

`= 100[(-4)^(1-1) + (-4)^(2-1) + (-4)^(3-1) + (-4)^(4-1)]`

`= 100[(-4)^(0) + (-4)^(1) + (-4)^(2) + (-4)^(3)]`

`= 100(1 - 4 + 16 - 64)`

`= 100(-51)`

`= color(blue)(-5100)`