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Please help me right away.

Please help me right away.-example-1
User Saul Uribe
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1 Answer

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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|<1, 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}


User Mivaweb
by
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