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Question Number 155802 by cortano last updated on 05/Oct/21

$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$$2\times 1!+4\times 2!+6\times 3!+...+200\times 100!$$

Answered by talminator2856791 last updated on 05/Oct/21

$$\mathrm{use}\mathrm{the}\mathrm{information}\mathrm{in}\mathrm{the}\mathrm{post}$$$$\mathrm{above}\mathrm{this}\mathrm{post}\mathrm{Q}.155803:$$$$\underset{k=1}{\sum ^n}k\times k!=(n+1)!-1$$$$2\times 1!+4\times 2!+.....+200\times 100!$$$$=2(1+1!+2\times 2!+.....+100\times 100!)$$$$=2\underset{k=1}{\sum ^{100}}k\times k!$$$$=2((100+1)!-1)$$$$=\overline{)\begin{array}{c}2\times 101!-2\end{array}}$$

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