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:

\(\displaystyle{\sum_{{{L}={0}}}^{{M}}}{L}^{{2}}{s}^{{L}}={0}^{{2}}+{1}^{{2}}{s}+{s}^{{2}}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}{''}+{s}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}'\)

\(\displaystyle{L}^{{2}}{s}^{{L}}={s}^{{2}}{L}{\left({L}-{1}\right)}{s}^{{{L}-{2}}}+{s}{L}{s}^{{{L}-{1}}}={s}^{{2}}{\left({s}^{{L}}\right)}{''}+{s}{\left({s}^{{L}}\right)}'\)

We get:

\(\displaystyle{\sum_{{{L}={0}}}^{{M}}}{L}^{{2}}{s}^{{L}}={0}^{{2}}+{1}^{{2}}{s}+{s}^{{2}}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}{''}+{s}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}'\)