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Question
1.1!+2.2!+3.3!+....+n.n!=
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Solution
The correct option is D (n+1)!−1
Given,
1.1!+2.2!+3.3!+........+n.n!=
Rearranging the terms, we get,
Sum=(2−1).1!+(3−1).2!+(4−1).3!...............+(n+1−1).n!
=2.1!−1.1!+3.2!−1.2!+4.3!−1.3!.............(n+1)n!−1n!
=2!−1!+3!−2!+4!−3!+................(n+1)!−n!
=(n+1)!−1
Given,
1.1!+2.2!+3.3!+........+n.n!=
Rearranging the terms, we get,
Sum=(2−1).1!+(3−1).2!+(4−1).3!...............+(n+1−1).n!
=2.1!−1.1!+3.2!−1.2!+4.3!−1.3!.............(n+1)n!−1n!
=2!−1!+3!−2!+4!−3!+................(n+1)!−n!
=(n+1)!−1
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