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I want to find

$$\sum\limits_{x=0}^{\infty} \frac{xe^{2x-5}5^x}{x!} $$

The book has this problem in says to use for the Taylor series of $e^x = \sum\limits_{j=1}^{\infty} \frac{x^j}{j!} $ . I then got no idea how the book came to a final answer of $4e^2$. Any help from anyone would be greatly appreciated :)

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    $\begingroup$ I'm not seeing any $i$'s in the sum. Is $x$ supposed to be $i$? $\endgroup$
    – Alex R.
    May 12 '17 at 16:55
  • $\begingroup$ There's a messy confusion between the running index, the variable and etc.....Please correct that. $\endgroup$
    – DonAntonio
    May 12 '17 at 16:58
  • $\begingroup$ Yeah my apology for the typo. Now it's edited. $\endgroup$ May 12 '17 at 17:01
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Hint

1) First huge hint: change that awfully lookin running index $\;x\;$ for $\;k,n,m\;$ or something similar.

2) Use the series for the exponential function with

$$\sum_{k=0}^\infty\frac{k\,3^{2k-5}5^k}{k!}=\frac1{e^5}\sum_{k=1}\frac{(5e^2)^k}{(k-1)!}$$

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