$ = \,\,\frac{1}{6}\,\sum\limits_{k\, = \,1}^n {\left( {2{k^3}\, + \,3{k^2}\, + \,k} \right)} $
$ = \,\frac{1}{3}.\,{\left\{ {\frac{{n\,(n\, + \,1)}}{2}} \right\}^2}\, + \,\,\frac{1}{2}\,\frac{{n\,(n\, + \,1)\,(2n\, + \,1)}}{6}\,\, + \,\,\frac{1}{6}\,\frac{{n\,(n\, + \,1)}}{2}$
${\text{a = }}{{\text{n}}^{\text{4}}}$ નો અચળાંક $ = \,\,\frac{1}{3}.\,\frac{{\text{1}}}{{\text{4}}};$
$\,\,{\text{b}}\, = {\text{ }}{{\text{n}}^{\text{3}}}$ નો અચળાંક $\, = \,\,\frac{1}{3}.\,\frac{1}{2}\,\, + \,\,\frac{1}{6} \,$ વગેરે.
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જો $A_{k}=\sum_{i=0}^{9}\left(\begin{array}{l}9 \\ i\end{array}\right)\left[\begin{array}{c}12 \\ 12-k+i\end{array}\right]+\sum_{i=0}^{8}\left(\begin{array}{c}8 \\ i\end{array}\right)\left[\begin{array}{c}13 \\ 13-k+i\end{array}\right]$
અને $A_{4}-A_{3}=190 \mathrm{p}$ હોય તો $p$ ની કિમંત મેળવો.