I was wondering how to find the sum of the power series $$\sum_{k=0}^\infty {2^k{x^k}-2k}{x^k}$$
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Do you mean $\sum_{k=0}^{\infty}(n^2-2^n)x^n$? – Dietrich Burde Oct 25 '19 at 14:57
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@DietrichBurde yes – Alex.G Oct 25 '19 at 14:58
2 Answers
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For the first, you can use the technique here twice to get the $k^2$. We have $$\sum_{k=0}^\infty x^k=\frac 1{1-x}\\ \frac d{dx}\sum_{k=0}^\infty x^k=\frac d{dx}\frac 1{1-x}\\ \sum_{k=0}^\infty xkx^{k-1}=x\frac 1{(1-x)^2}\\ \frac d{dx}\sum_{k=0}^\infty xkx^{k-1}=\frac{1+x}{(1-x)^3}\\ \sum_{k=0}^\infty xk^2x^{k-1}=\frac{x+x^2}{(1-x)^3}$$
For the second, now that it has been updated, you can use the second line.
Ross Millikan
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and then do you just subtract the sum of the second from the sum of the first? @rossmillikan – Alex.G Oct 25 '19 at 15:33
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I had missed a minus sign in the Alpha output when I answered initially. Fixed. Can you subtract the fractions? – Ross Millikan Oct 25 '19 at 15:48
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yes but it does not give the correct answer. the right answer is $$\frac{3x^2-x}{{(1-x)}^3}$$ @rossmillikan – Alex.G Oct 25 '19 at 15:51
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When I answered the second term was $2^kx^k$. Now you have $2kx^k$. That is a different problem. You can take the result in the middle of my answer to get the second term. – Ross Millikan Oct 25 '19 at 15:53
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If $|z|<1$, then $$\sum_{k=0}^{\infty} z^k =\frac{1}{1-z} ~~~(1)$$ D.w.r.t. $z$ $$\sum_{k=1}^{\infty} k z^k =\frac{z}{(1-z)^2}~~~~(2).$$ So $$S=\sum_{k=0}^{\infty} [(2x)^k -2 k x^k]$$ By (1) and (2) we get $$S=\frac{1}{1-2x}-\frac{2x}{(1-x)^2}=\frac{5x^2-4x+1}{(1-2x)(1-x)^2}, ~if~ |x|<\frac{1}{2}.$$
Z Ahmed
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