Question

1(1!)+2(2!)+-------------2012(2012!)=s find s?

Answer 1

\[\sum_{k=1}^{n}k\cdot k!\]

\[\sum_{k=1}^{n}(k+1-1)\cdot k!\]

\[\sum_{k=1}^{n}((k+1)-1)\cdot k!\]
\[=\sum_{k=1}^{n}((k+1)k!-k!)\]
\[=\sum_{k=1}^{n}(k+1)k!-\sum_{k=1}^{n}k!\]
\[=\sum_{k=1}^{n}(k+1)!-\sum_{k=1}^{n}k!\]

\[=(n+1)!+n!+(n-1)!+\cdots+2!-(n!+(n-1)!+(n-2)!+\cdots+2!+1!\]

\[=(n+1)!-n!+n!-(n-1)!+(n-1)!+\cdots-3!+3!-2!+2!-1!\]

\[=(n+1)!-1\]