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Sum of the series is
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a
(n+1)!
b
(n+2)!−1
c
n(n+1)!
d
none of these
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Correct option is C
We can writer2+1=(r+2)(r+1)−3(r+1)+2Thus ∑r=1n r2+1(r!)=∑r=1n [(r+2)(r+1)−(r+1)−2{(r+1)−1}]r! =∑r=1n [(r+2)!−(r+1)!]−2∑r=1n {(r+1)!−r!}=(n+2)!−2!−2{(n+1)!−1}=n(n+1)!Similar Questions
Let (where, [ . ] denotes greatest integer function) is equal to
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