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What is the appropriate way to simplify such an expression. i am unsure of how to use the series i know to apply to this situation

$\sum_{L=0}^{M}s^{L}L^{2}$

do i modify such a series as power series, or is there a more efficient series to use here?

thank you very much!!

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    @Mohamed Thanks - I figured OP would want a calculus-only kind of answer.2012-06-19

3 Answers 3

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Rewrite $L^2 = L(L-1)+L. $ Then,

$\begin{align} \sum_{L=0}^{M} { L^2 s^L } &= \sum_{L=0}^{M} {L(L-1)s^L} + \sum_{L=0}^{M} {Ls^L}\\ &=s^2 \cdot \frac{\partial^2}{\partial s^2} \sum_{L=0}^{M}{s^L} + s \cdot \frac{\partial}{\partial s}\sum_{L=0}^{M} {s^L}\\ \end{align}$

You can find formulas for the summations with detailed descriptions of their derivations by searching for "Geometric Progression."

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Write : $L^2=L(L-1)+L$ and use derivative. For $ L \geq 2$ : $L^2 s^L = s^2L(L-1)s^{L-2}+ s Ls^{L-1}= s^2(s^L)'' + s (s^L)'$ We get : $\sum_{L=0}^M L^2s^L=0^2+1^2 s + s^2 \left(\sum_{L=2}^{M} s^L \right)''+s \left(\sum_{L=2}^{M} s^L \right)'$

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    @ alex.jordan Thank you very much.I mean :$0^2 s^0 + 1^2 s^1$2012-07-13
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Try to make the inner expression look like a derivative: $ \begin{align} \sum_{L=0}^M\left(Ls^{L-1}\right)sL & =s\sum_{L=0}^M\left(\partial_ss^L\right)L\\ & =s\partial_s\sum_{L=0}^Ms^LL\\ & =s\partial_s\sum_{L=0}^M\left(Ls^{L-1}\right)s\\ & =s\partial_s\left(s\sum_{L=0}^M\left(Ls^{L-1}\right)\right)\\ & =s\partial_s\left(s\sum_{L=0}^M\partial_ss^L\right)\\ & =s\partial_s\left(s\partial_s\sum_{L=0}^Ms^L\right)\\ & =s\partial_s\left(s\partial_s\frac{s^{M+1}-1}{s-1}\right)\\ \end{align}$ Now just take it from here, simplifying from the inside out.