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Question

The sum of the series (12+1).1!+(22+1).2!+(32+1).3!+....+(n2+1).n! is:

A
(n + 1). (n+2)!
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B
n.(n+1)!
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C
(n + 1). (n+1)!
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D
none of these
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Solution

The correct option is B n.(n+1)!
Tn = [ n (n + 1) - (n - 1) ] n! = n.(n+1)! - (n - 1).n!
Now put n = 1, 2, 3, ...... , n and add

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