# What is the appropriate way to simplify such an expression.

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

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aquariump9
Write : $$\displaystyle{L}^{{2}}={L}{\left({L}-{1}\right)}+{L}$$ and use derivative. For $$\displaystyle{L}\geq{2}$$:
$$\displaystyle{L}^{{2}}{s}^{{L}}={s}^{{2}}{L}{\left({L}-{1}\right)}{s}^{{{L}-{2}}}+{s}{L}{s}^{{{L}-{1}}}={s}^{{2}}{\left({s}^{{L}}\right)}{''}+{s}{\left({s}^{{L}}\right)}'$$
We get:
$$\displaystyle{\sum_{{{L}={0}}}^{{M}}}{L}^{{2}}{s}^{{L}}={0}^{{2}}+{1}^{{2}}{s}+{s}^{{2}}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}{''}+{s}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}'$$
###### Not exactly what youâ€™re looking for?
Neunassauk8
Try to make the inner expression look like a derivative:
$$\displaystyle{\sum_{{{L}={0}}}^{{M}}}{\left({L}{s}^{{{L}-{1}}}\right)}{s}{L}={s}{\sum_{{{L}={0}}}^{{M}}}{\left({d}_{{s}}{s}^{{L}}\right)}{L}$$
$$\displaystyle={s}{d}_{{s}}{\sum_{{{L}={0}}}^{{M}}}{s}^{{L}}{L}$$
$$\displaystyle={s}{d}_{{s}}{\sum_{{{L}={0}}}^{{M}}}{\left({L}{s}^{{{L}-{1}}}\right)}{s}$$
$$\displaystyle={s}{d}_{{s}}{\left({s}{\sum_{{{L}={0}}}^{{M}}}{\left({L}{s}^{{{L}-{1}}}\right)}\right)}$$
$$\displaystyle={s}{d}_{{s}}{\left({s}{\sum_{{{L}={0}}}^{{M}}}{d}_{{s}}{s}^{{L}}\right)}$$
$$\displaystyle={s}{d}_{{s}}{\left({s}{d}_{{s}}{\sum_{{{L}={0}}}^{{M}}}{s}^{{L}}\right)}$$
$$\displaystyle={s}{d}_{{s}}{\left({s}{d}_{{s}}{\frac{{{s}^{{{M}+{1}}}-{1}}}{{{s}-{1}}}}\right)}$$
Now just take it from here, simplifying from the inside out.
karton

Rewrite $$L^2=L(L-1)+L$$. Then,
$$\sum_{L=0}^M L^2s^L=\sum_{L=0}^M L(L-1)s^L+\sum_{L=0}^M Ls^L$$
$$=s^2\cdot\frac{d^2}{ds^2}\sum_{L=0}^M s^L+s\cdot\frac{d}{ds}\sum_{L=0}^M s^L$$