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I am find the expected value for something and I have simplified it to the below, but I am not sure how to further solve it. Can soembody help and see if I am on the right track?

My working: https://i.stack.imgur.com/Ftwo1.png

I converted the 1 + 2 + 3... part to an AP, and the (1/4)^2k-1 to a GP, and combined them together. Does it work like that?

Iva
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 12 '21 at 01:57
  • Hint: https://math.stackexchange.com/questions/333192/solve-sum-nxn (if computing $\sum_{k=1}^\infty k(1/4)^{2k-1}(3/4)$ is your goal). And no what you wrote does not make sense. – justadzr Sep 12 '21 at 01:58
  • Better to write out the whole question here, than to make users chase a link offsite. – Gerry Myerson Sep 12 '21 at 03:33
  • Hint: The given series is an AGP, you may visit here to know how to find the sum of an infinite AGP. –  Sep 12 '21 at 22:16

1 Answers1

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You are trying to compute $S = \sum_{k=1}^\infty k a^{2k-1}$.

You can write: $S = a + 2a^3 + 3a^5 + 4a^7\ldots$

Also,

$a^2S = a^3 + 2a^5 + 3a^7 + 4a^9 \ldots$

Now compute the difference between the two equations to get:

$S(1-a^2) = a + a^3 + a^5 + \ldots = \frac{a}{1-a^2}$

implying

$S = \frac{a}{(1-a^2)^2}$.

You can plug in $a$ to get the desired result.

Godfather
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