A) \[(n+1)!\]
B) \[(n+1)!+1\]
C) \[(n+1)!-1\]
D) None of these
Correct Answer: C
Solution :
We have, \[1.1!+2.2!+3.3!+...+n.n!\] \[=\sum\limits_{r=1}^{n}{r}.(r!)=\sum\limits_{r=1}^{n}{[(r+1)r!-r!]}\] \[=\sum\limits_{r=1}^{n}{[(r+1)!-r!]}\] \[=(2!-1!)+(3!-2!)+....+[(n+1)!-n!]\] \[=(n+1)!-1!=(n+1)!-1\]
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