Q.
The sum of n terms of 1 . 2 . 3+2 . 3 . 4+....... will be
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a
n(n+1)(n+2)(n+3)4
b
2n(n+1)(n+2)(n+3)3
c
(n+2)(n+1)(n+3)4
d
n(n−1)(n−2)(n−3)4
answer is A.
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Detailed Solution
We have, Tn=n(n+1)(n+2)=n3+3n2+2n Sum of n terms =Sn=∑n3+3∑n2+2∑n =14n2(n+1)2+36n(n+1)(2n+1)+22n(n+1) =nn+12nn+12+2n+1+2 =14n(n+1)(n2+5n+6) =14n(n+1)(n+2)(n+3)
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