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Question Number 155842 by zainaltanjung last updated on 05/Oct/21 | ||
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$$\mathrm{Prof}\:\mathrm{that}: \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\mathrm{n}.\mathrm{n}!=\mathrm{11}!−\mathrm{1} \\ $$ | ||
Commented by puissant last updated on 05/Oct/21 | ||
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$${Q}\mathrm{155803} \\ $$ | ||
Answered by mr W last updated on 05/Oct/21 | ||
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$$\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}{n}×{n}! \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\left[\left({n}+\mathrm{1}\right)×{n}!−{n}!\right] \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\left[\left({n}+\mathrm{1}\right)!−{n}!\right] \\ $$$$=\left({k}+\mathrm{1}\right)!−\mathrm{1}! \\ $$$$=\left({k}+\mathrm{1}\right)!−\mathrm{1} \\ $$ | ||