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Question
1+1.1!+2.2!+3.3!+...+n.n! is equal to
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Solution
The correct option is C (n+1)!
1+1.1!+2.2!+3.3!+...+n.n!=1+(2.1!−1!)+(3.2!−2!)+(4.3!−3!)+.....((n+1).n!−n!)=1+(2.1!+3.2!+4.3!+....+(n+1).n!)−(1!+2!+3!+...+n!)=1+(2!+3!+4!+.....+(n+1)!)−(1!+2!+3!+...+n!)=1+(n+1)!−1=(n+1)!
1+1.1!+2.2!+3.3!+...+n.n!
=1+(2.1!−1!)+(3.2!−2!)+(4.3!−3!)+.....((n+1).n!−n!)
=1+(2.1!+3.2!+4.3!+....+(n+1).n!)−(1!+2!+3!+...+n!)
=1+(2!+3!+4!+.....+(n+1)!)−(1!+2!+3!+...+n!)
=1+(n+1)!−1=(n+1)!
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