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If 1+5+12+22+35+ to n terms =n2(n+1)2 nth term of series is

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a
n(4n−1)3
b
n(3n−1)2
c
n(3n+1)2
d
n(4n+1)3

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detailed solution

Correct option is B

Let P(n):1+5+12+22+35+…+(n terms ) =n2(n+1)2nth term of LHS = P(n) - P(n - 1) ⇒ P(n)−P(n−1)=n2(n+1)2−(n−1)2n2⇒ P(n)−P(n−1)=n2n2+n−n2+2n−1∴ Tn=P(n)−P(n−1)=n2(3n−1)


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