The sum of the series 12+(12+22)+(12+22+32) to n terms = ...

Question

The sum of the series 12+(12+22)+(12+22+32) to n terms = ... (A) (n(n+1)2(n+2)/12) (B) (n(n+1)2(n+3)/12) (C) (n(n+2)2(n+1)/12) (D) (n(n+2)2(n+1)/14)

Tardigrade Admin · Accepted Answer

Tn = (12+22+32 +...+n2) = (n(n+1)(2n+1)/6) = (1/6)[2n3+3n2 +n] ∴ Sn = ∑ limitsk=1n Tk = (1/6)[2∑ limits k=1n k3 +3∑ limits k=1n k2 +∑ limits k=1n k] = (1/6)[2 (n(n+1)2/2) 2 + (3n(n+1)(2n+1)/6) + (n(n+1)/2)] =(1/6) [(n2(n+1)2/2) +(n(n+1)(2n+1)/2) +(n(n+1)/2)] = (n(n+1)/12)[(n(n+1)+2n+1+1/1)] = (n(n+1)2(n+2)/12)

Q. The sum of the series $1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2})$ to $n$ terms = ...

$\frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}$

$\frac{n ( n + 1 ) ^{2} ( n + 3 )}{12}$

$\frac{n ( n + 2 ) ^{2} ( n + 1 )}{12}$

$\frac{n ( n + 2 ) ^{2} ( n + 1 )}{14}$

Q. The sum of the series 12+(12+22)+(12+22+32) to n terms = ...

12n(n+1)2(n+2)​

12n(n+1)2(n+3)​

12n(n+2)2(n+1)​

14n(n+2)2(n+1)​

Tn​=(12+22+32+...+n2) =6n(n+1)(2n+1)​ =61​[2n3+3n2+n] ∴Sn​=k=1∑n​Tk​ =61​[2k=1∑n​k3+3k=1∑n​k2+k=1∑n​k] =61​[2{2n(n+1)2​}2+63n(n+1)(2n+1)​+2n(n+1)​] =61​[2n2(n+1)2​+2n(n+1)(2n+1)​+2n(n+1)​] =12n(n+1)​[1n(n+1)+2n+1+1​] =12n(n+1)2(n+2)​

Q. The sum of the series $1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2})$ to $n$ terms = ...

$\frac{n ( n + 1 ) ^{2} ( n + 2 )}{12}$

$\frac{n ( n + 1 ) ^{2} ( n + 3 )}{12}$

$\frac{n ( n + 2 ) ^{2} ( n + 1 )}{12}$

$\frac{n ( n + 2 ) ^{2} ( n + 1 )}{14}$