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[lkdash]

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

 

[pka]

You need to know that if \(|r|<1\) then \(\displaystyle\sum\limits_{n = J}^\infty {{r^n} = \frac{{{r^J}}}{{1 - r}}} \)

 

[Steven G]

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

 

[HallsofIvy]

The "hard part", of course, is that infinite sum.

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

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"!

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

Add 2 to that.