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lkdash

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Jul 15, 2020
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I feel the Answer is A and E. Can you guys tell me which is correct?

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You need to know that if r<1|r|<1 then n=Jrn=rJ1r\displaystyle\sum\limits_{n = J}^\infty {{r^n} = \frac{{{r^J}}}{{1 - r}}}
 
How can the answer be A and E??!! Might you mean A or E? In math you must know the difference between and and or
 
The "hard part", of course, is that infinite sum.

Let S=15+152+153++++15n+\displaystyle S= \frac{1}{5}+ \frac{1}{5^2}+ \frac{1}{5^3}+ \cdot+ \cdot+ \cdot+ \frac{1}{5^n}+ \cdot\cdot\cdot
Factor out 15\displaystyle \frac{1}{5}:
S=15(1+15+152++++15n1+)\displaystyle S= \frac{1}{5}\left(1+ \frac{1}{5}+ \frac{1}{5^2}+ \cdot+ \cdot+ \cdot+ \frac{1}{5^{n-1}}+ \cdot\cdot\cdot\right).

Do you see that, after the "1+ " the sum on the right is just the original sum, with the index shifted? And since the sum is infinite, that is just the original "S"!

S=15(1+S)\displaystyle S= \frac{1}{5}(1+ S).
So 5S=1+S\displaystyle 5S= 1+ S, 4S=1\displaystyle 4S= 1, and S=14\displaystyle S= \frac{1}{4}.

Add 2 to that.
 
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