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Prove that the sum of the infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ is 23.

My approach I got the following term $S_n=\sum_1^\infty\frac{4n^2}{2^n}-\sum_1^\infty\frac{1}{2^n}$.

For $\sum_1^\infty\frac{1}{2^n}$ the answer is 1 as it forms a geometric series but I am bot able to find the solution to $\sum_1^\infty\frac{4n^2}{2^n}$.

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    $\begingroup$ Do you mean $1.3$ (a decimal) or $1\cdot 3$ (a product)? $\endgroup$
    – quarague
    Mar 28, 2019 at 13:29

2 Answers 2

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$$\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+…=\sum_{n=1}^{\infty }\frac{(2n-1)(2n+1)}{2^n}=\sum_{n=1}^{\infty }\frac{(4n^2-1)}{2^n}$$ depending on the geometric series $$\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$ $$(\frac{1}{1-x})'=\sum_{n=1}^{\infty }nx^{n-1}$$ $$x(\frac{1}{1-x})'=\sum_{n=1}^{\infty }nx^{n}$$ $$(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }n^2x^{n-1}$$ $$x(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }n^2x^{n}$$ $$4x(x(\frac{1}{1-x})')'=\sum_{n=1}^{\infty }4n^2x^{n}$$ $$4x(x(\frac{1}{1-x})')'-\frac{1}{1-x}=\sum_{n=1}^{\infty }4n^2x^{n}-\sum_{n=0}^{\infty }x^n$$ $$4x(x(\frac{1}{1-x})')'-\frac{1}{1-x}+1=\sum_{n=1}^{\infty }4n^2x^{n}-\sum_{n=1}^{\infty }x^n$$ so $$\sum_{n=1}^{\infty }x^n(4n^2-1)=\frac{4x^2+4x}{(1-x)^3}-\frac{x}{1-x}$$ now let $x=1/2$ to get $23$

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Hint. One may used this $$\sum_{n\ge1}n^{2}x^n=\frac{x}{(1-x)^{2}}+\frac{2x^{2}}{(1-x)^{3}},\qquad |x|<1,$$ proved for example here.

Hope you can take from here.

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