What is the sum of the infinite series 1^2 + | vedclass

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Question

(A) Let the sum be $S = 1^2 + 2^2 x + 3^2 x^2 + \dots$. We know that the sum of the infinite geometric series is $\sum_{n=0}^{\infty} x^n = \frac{1}{1

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$(1 + x) / (1 - x)^3$

Vedclass · Answer

(A) Let the sum be $S = 1^2 + 2^2 x + 3^2 x^2 + \dots$. We know that the sum of the infinite geometric series is $\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x| Differentiating both sides with respect to $x$,we get $\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}$. Multiplying by $x$,we get $\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}$. Differentiating again with respect to $x$,we get $\sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{d}{dx} \left( \frac{x}{(1-x)^2} \right) = \frac{(1-x)^2(1) - x(2(1-x)(-1))}{(1-x)^4} = \frac{(1-x) + 2x}{(1-x)^3} = \frac{1+x}{(1-x)^3}$. Thus,the sum $S = 1^2 + 2^2 x + 3^2 x^2 + \dots = \sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{1+x}{(1-x)^3}$.

What is the sum of the infinite series $1^2 + 2^2 x + 3^2 x^2 + \dots$?

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