My approach : I tried using integral calculus and using infinite geometric series..however it didn't match..any trick?
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$\begingroup$Hint: $$\frac{1}{1-x}=1+x+x^2+x^3+\dots$$ Differentiating with respect to $x$ and assuming $|x|<1$, which guarantees uniform convergence:
$$\dfrac{d}{dx}\left(\frac{1}{1-x}\right)=1+2x+3x^2+\dots.$$ (you should also check the radius of convergence of the resulting expression.)
An Alternative approach is based on the assumption of absolute convergence of the series:
$$S = 1 + x + x^2 + \dots$$ $$ + x + x^2 + \dots$$ $$ + x^2 + \dots$$ $$ \dots $$
$\endgroup$ $\begingroup$F = 1 + x + x^2 + ...
Your sum is F + x*F + x^2*F + x^3 * F ...
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