Question

Please help me right away.

Answer 1

Two quick changes before diving into action:

First of all, we can factor the 100 out of the sum:

$\sum_{k=1}^\infty 100(-0.9)^{k-1} = 100\sum_{k=1}^\infty (-0.9)^{k-1}$

Secondly, we can see that the index k starts from 1, but the exponent is k-1. This means that when k=1, the exponent is actually 0. When k=2, the exponent is actually 1, and so on. So, we can rewrite the sum as

$100\sum_{k=1}^\infty (-0.9)^{k-1} = 100\sum_{k=0}^\infty (-0.9)^k$

Now, let's focus on the sum. First of all, it converges, because every sum like

$\sum_{k=0}^\infty a^k$

converges if and only if $|a|$, which is the case because

$|-0.9| = 0.9 < 1$

Also, in this case we have

$\sum_{k=0}^\infty a^k = \cfrac{1}{1-a}$

so in your case you have

$\sum_{k=0}^\infty (-0.9)^k = \cfrac{1}{1-(-0.9)} = \cfrac{1}{1.9}$

and let's not forget the 100 we factored at the beginning!

$100\sum_{k=0}^\infty (-0.9)^k= 100\cdot\cfrac{1}{1.9} = \cfrac{100}{1.9}$