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Question Number 109169 by ZiYangLee last updated on 21/Aug/20
The value of 1∙1! + 2∙2! + 3∙3! +...+n∙n! is
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{1}\centerdot\mathrm{1}!\:+\:\mathrm{2}\centerdot\mathrm{2}!\:+\:\mathrm{3}\centerdot\mathrm{3}!\:+…+{n}\centerdot{n}! \\ $$$$\mathrm{is}\: \\ $$
Answered by Dwaipayan Shikari last updated on 21/Aug/20
Σ_(n=1) ^n n.n!=Σ_(n=1) ^n (n+1)n!−n!=2.1!−1!+3.2!−2!+4.3!−3!+.. =2!−1!+3!−2!+4!−3!+....+(n+1)!−n! =(n+1)!−1 Or Σ_(n=1) ^n n.n!=Σ_(n=1) ^n (n+1)n!−n!=Σ_(n=1) ^n (n+1)!−n!=2!−1!+3!−2!+....=(n+1)!−1
$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{n}.{n}!=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}+\mathrm{1}\right){n}!−{n}!=\mathrm{2}.\mathrm{1}!−\mathrm{1}!+\mathrm{3}.\mathrm{2}!−\mathrm{2}!+\mathrm{4}.\mathrm{3}!−\mathrm{3}!+.. \\ $$$$=\mathrm{2}!−\mathrm{1}!+\mathrm{3}!−\mathrm{2}!+\mathrm{4}!−\mathrm{3}!+….+\left({n}+\mathrm{1}\right)!−{n}! \\ $$$$=\left({n}+\mathrm{1}\right)!−\mathrm{1} \\ $$$${Or} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{n}.{n}!=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}+\mathrm{1}\right){n}!−{n}!=\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}\left({n}+\mathrm{1}\right)!−{n}!=\mathrm{2}!−\mathrm{1}!+\mathrm{3}!−\mathrm{2}!+….=\left({n}+\mathrm{1}\right)!−\mathrm{1} \\ $$
Commented by JDamian last updated on 22/Aug/20
Please, correct your notation mistake.

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