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What series is this? Is there a formula? Also, is there a quick way to calculate $E(X^2)$?

$E(X) = \sum _{k=1}^{\infty \:}\left(\frac{1}{2}\right)^k$*k

I've tried using
$E(X) = 0.5+0.5^2+0.5^3 + ...$
$ 0.5 E(X) = 0.5^2+0.5^3 * 2 + 0.5^4 *3 + ...$
$ 0.5 E(X) = 0.5 + 0.5^2 + 0.5^3 + ... $

The third line is the difference of the first two lines here. Using geometric series, the third line gives 0.5 E(X) = 1, so E(X) = 2, but is there a faster way?

Thank you!

tee
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  • sorry by formula I meant the E(X) = summation ... that line. I also realised there was a typo in the E(x) line. can u pls have a look at it again? ty! – tee Apr 09 '23 at 23:20
  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Apr 09 '23 at 23:22
  • @JeanMarie yep thank u sm!!! – tee Apr 09 '23 at 23:35
  • @Shaun thanks for the feedback!! I've edited it now – tee Apr 09 '23 at 23:36

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