Find formula for f[x] whose expansion in powers of x is

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I am trying to find formula for f[x] whose expansion in powers of x is:
x + 2^2 x^2 + 3^2 x^3 + 4^2 x^4 + ... + k^2 x^k + ...

I know that this is a variation on x/(1-x). I also know that x/(1-x)^2 yields expansion in power of x = kx^k. I cannot figure out the manipulation which would give the expansion k^2x^k.

Any help is appreciated
 
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eclayj said:
I am trying to find formula for f[x] whose expansion in powers of x is:
x + 2^2 x^2 + 3^2 x^3 + 4^2 x^4 + ... + k^2 x^k + ...

I know that this is a variation on x/(1-x). I also know that x/(1-x)^2 yields expansion in power of x = kx^k. I cannot figure out the manipulation which would give the expansion k^2x^k.

Any help is appreciated

Well, if I had a series expansion that gave me terms kxk and I differentiated it that would give k2xk-1 and if I multiplied by x ...
 
Thank you very much. I need more practice and hopefully I will get better at seeing where a formula can be differentiated to get you close.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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