William Cleghorn

Answered

2022-01-06

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

$\sum _{L=0}^{M}{s}^{L}{L}^{2}$

Answer & Explanation

aquariump9

Expert

2022-01-07Added 40 answers

Write : ${L}^{2}=L(L-1)+L$ and use derivative. For $L\ge 2$ :

$L}^{2}{s}^{L}={s}^{2}L(L-1){s}^{L-2}+sL{s}^{L-1}={s}^{2}\left({s}^{L}\right){}^{\u2033}+s{\left({s}^{L}\right)}^{\prime$

We get:

$\sum _{L=0}^{M}{L}^{2}{s}^{L}={0}^{2}+{1}^{2}s+{s}^{2}\left(\sum _{L=2}^{M}{s}^{L}\right){}^{\u2033}+s{\left(\sum _{L=2}^{M}{s}^{L}\right)}^{\prime}$

We get:

Neunassauk8

Expert

2022-01-08Added 30 answers

Try to make the inner expression look like a derivative:

$\sum _{L=0}^{M}\left(L{s}^{L-1}\right)sL=s\sum _{L=0}^{M}\left({d}_{s}{s}^{L}\right)L$

$=s{d}_{s}\sum _{L=0}^{M}{s}^{L}L$

$=s{d}_{s}\sum _{L=0}^{M}\left(L{s}^{L-1}\right)s$

$=s{d}_{s}\left(s\sum _{L=0}^{M}\left(L{s}^{L-1}\right)\right)$

$=s{d}_{s}\left(s\sum _{L=0}^{M}{d}_{s}{s}^{L}\right)$

$=s{d}_{s}\left(s{d}_{s}\sum _{L=0}^{M}{s}^{L}\right)$

$=s{d}_{s}\left(s{d}_{s}\frac{{s}^{M+1}-1}{s-1}\right)$

Now just take it from here, simplifying from the inside out.

Now just take it from here, simplifying from the inside out.

karton

Expert

2022-01-11Added 439 answers

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