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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

Thevalueof1⋅1!+2⋅2!+3⋅3!+...+n⋅n!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

∑nn=1n.n!=∑nn=1(n+1)n!−n!=2.1!−1!+3.2!−2!+4.3!−3!+..=2!−1!+3!−2!+4!−3!+....+(n+1)!−n!=(n+1)!−1Or∑nn=1n.n!=∑nn=1(n+1)n!−n!=∑nn=1(n+1)!−n!=2!−1!+3!−2!+....=(n+1)!−1

Commented by JDamian last updated on 22/Aug/20

Please, correct your notation mistake.

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