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3 years ago

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Please answer question. I need some math to explain the answer, not just the answer. Ignore my work, that was wrong. Please help

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WILL GIVE BRAINLISTSSSSS!!!!

1 answer:

Virty [35]3 years ago

3 0

The correct answer to your question is number 2

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Please help me solve this

Viktor [21]

You can factor the 32 out of the sum:

$\displaystyle \sum_{m=1}^\infty 32 \cdot \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1}$

We can also change the index as follows

$\displaystyle 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m}$

Now, we have a theorem that states that the series

$\displaystyle \sum_{m=1}^\infty a^m$

converges if and only if $|a|$ , and in this case we have

$\displaystyle \sum_{m=1}^\infty a^m = \dfrac{1}{1-a}$

This is your case, because you have

$|a|=\dfrac{1}{2}$

which implies that your series converges, and the value is

$\displaystyle 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m} = 32 \cdot \dfrac{1}{1-\frac{1}{2}} = 32\cdot 2 = 64$

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|-11| > |-10|
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