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Question

Sum of the series nr=1(r2+1)r! is ______

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

The correct option is C n(n+1)!

nr=1(r2+1)r!


nr=1(r2+1+2r2r)r!


nr=1((r+1)2.r!2r.r!)


nr=1((r+1).(r+1)!2r.r!)


nr=1[(r+21)(r+1)!2((r+11)r!)]


nr=1[(r+2)(r+1)!(r+1)!2((r+1)r!r!)]


nr=1[(r+2)!(r+1)!2.(r+1)!+2.r!)]


T1=3!2!2.2!+2.1!


T2=4!3!2.3!+2.2!


T3=5!4!2.4!+2.3!

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Tn=(n+2)!(n+1)!2.(n+1)!+2.n!

Sum=Sn=(n+2)!2!2(n+1)!+2.1!


Sn=(n+2)!2(n+1)!2+2


Sn=(n+1)![n+22]


Sn=n.(n+1)!


Correct answer is C


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