The sum of the series $3\times 1^2+5\times 2^2+7\times 3^2+\dots$ is

detailed solution

Correct option is B

Let given series is 5 = 3 x 12 + 5 x 22 + 7 x 32 + ...First, we will split the given series into two parts which are 3, 5,7, ... and 12,22,32, , . . and find the nth term of each partseparately to find the nth term of the given series. Tn = (rth term of 3,5, 7,. . ) x (nth term of 1,2,3, . . )2=[3+(n−1)2][1+(n−1)]2=(3+2n−2)(n)2=(2n+1)n2=2n3+n2 Now, S=ΣTn=2Σn3+Σn2=2n(n+1)22+n(n+1)(2n+1)6∵Σn3=n(n+1)22=n(n+1)22×n(n+1)2+2n+13=n(n+1)23n(n+1)+2n+13=n(n+1)6×3n2+3n+2n+1=n(n+1)3n2+5n+16