You are watching: Determine the sum of the following series.
Which an approach should I use to prove the the sum is 6?
It is equal to $f(x)=\sum_n \geq 0 n^2 x^n$ evaluate at $x=1/2$.To compute this function of $x$, compose $n^2 = (n+1)(n+2)-3(n+1)+1$, so the $f(x)=a(x)+b(x)+c(x)$ with:
$a(x)= \sum_n \geq 0 (n+1)(n+2) x^n = \fracd^2dx^2 \left( \sum_n \geq 0 x^n\right) = \frac2(1-x)^3$
$b(x)=\sum_n \geq 0 3 (n+1) x^n = 3\fracddx \left( \sum_n \geq 0 x^n \right) = \frac3(1-x)^2$
$c(x)= \sum_n \geq 0 x^n = \frac11-x$
So $f(1/2)=\frac2(1/2)^3-\frac3(1/2)^2 + \frac11/2 = 16-12+2=6$.
The "technique" is to include a parameter in the series, to do the multiplication through $n+1$ appear as differentiation.
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$\begingroup$ +1 for presenting the as an instance of a basic technique, quite than a trick that would show up to occupational only because that this case. An additional general technique, i m sorry is probably equivalent to yours, would be together follows. Create the similar-looking unknown integral $\int x^2 2^-x dx$. Combine it by parts. Note that the an outcome has the form $2^-xP(x)$, wherein $P$ is a polynomial. Conjecture that the partial sum in the discrete case has the kind $S_n=2^-nP(n)$. Compose $S_n-S_n-1=n^22^-n$, and determine $P$. $\endgroup$
For $x$ in a ar of $1$, let$$f(x) = \sum\limits_n = 0^\infty \fracx^n 2^n = \sum\limits_n = 0^\infty \bigg(\fracx2\bigg)^n = \frac11 - x/2 = \frac22 - x.$$Thus, top top the one hand,$$f'(x) = \sum\limits_n = 0^\infty \fracnx^n - 1 2^n \;\; \rm and \;\; f''(x) = \sum\limits_n = 0^\infty \fracn(n - 1)x^n - 2 2^n ,$$and, top top the other hand,$$f'(x) = \frac2(2 - x)^2 \;\; \rm and \;\; f''(x) = \frac4(2 - x)^3 .$$Hence,$$f'(1) = \sum\limits_n = 0^\infty \fracn2^n = 2 \;\; \rm and \;\; f''(1) = \sum\limits_n = 0^\infty \fracn(n - 1)2^n = 4.$$Finally, $$\sum\limits_n = 0^\infty \fracn^2 2^n = f'(1) + f''(1) = 6.$$The idea here was to think about the Probability-generating role of the geometric$(1/2)$ distribution.