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Sum of the series \mathrm{\sum_{r=1}^{n}\left(r^{2}+1\right)(r !)} is

Option: 1

\mathrm{(n+1) !}


Option: 2

\mathrm{(n+2)!-1}


Option: 3

\mathrm{n(n+1) !}


Option: 4

none of these


Answers (1)

best_answer

We can write

\mathrm{r^{2}+1=(r+2)(r+1)-3(r+1)+2}

Thus,
\mathrm{\sum_{r=1}^{n}\left(r^{2}+1\right)(r !)=\sum_{r=1}^{n}[(r+2)(r+1)-(r+1)-2\{(r+1)-1\}] r !}

\begin{gathered} \\ =\sum_{r=1}^{n}[(r+2) !-(r+1) !]-2 \sum_{r=1}^{n}\{(r+1) !-r !\} \\ =(n+2) !-2 !-2\{(n+1) !-1\}=n(n+1) ! \end{gathered}

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Gunjita

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