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Question Number 155803 by cortano last updated on 05/Oct/21

$$\mathrm{proof}\mathrm{that}$$$$\underset{\mathrm{n}=1}{\sum ^{10}}\mathrm{n}\times \mathrm{n}!=11!-1$$

Answered by som(math1967) last updated on 05/Oct/21

$$1\times 1!+2\times 2!+3\times 3!+...+10\times 10!+1-1$$$$=2+2\times 2!+3\times 3!+...+10\times 10!-1$$$$=2!(1+2)+3\times 3!+4\times 4!+...+10\times 10!-1$$$$=3!+3\times 3!+4\times 4!+...+10\times 10!-1$$$$=3!(1+3)+4\times 4!+...+10\times 10!-1$$$$=4!(1+4)+5\times 5!+...+10\times 10!-1$$$$=5!(1+5)+...+10\times 10!-1$$$$=...$$$$=10!(1+10)-1=11!-1\left[proved\right]$$

Answered by puissant last updated on 05/Oct/21

$$\underset{n=1}{\sum ^{10}}n\times n!=\underset{n=1}{\sum ^{10}}(n+1-1)\times n!$$$$=\underset{n=1}{\sum ^{10}}(n+1)\times n!-n!=\underset{n=1}{\sum ^{10}}\{(n+1)!-n!\}$$$$=2!-1!+3!-2!+4!-3!+....+11!-10!$$$$=11!-1!$$$$$$$$\therefore \because \underset{n=1}{\sum ^{10}}n\times n!=11!-1..$$

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