The sum of n terms of 12+(12+22)+(12+22+32) + ldots is-

Question

The sum of n terms of 12+(12+22)+(12+22+32) + ldots is- (A) ( n ( n +1)(2 n +1)/6) (B) ( n ( n +1)(2 n -1)/6) (C) (1/12) n(n+1)2(n+2) (D) (1/12) n2(n+1)2

Tardigrade Admin · Accepted Answer

Correct answer is (c) (1/12) n(n+1)2(n+2)

Q. The sum of $n$ terms of $1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2})$ $+ \dots$ is-

$\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

$\frac{n ( n + 1 ) ( 2 n - 1 )}{6}$

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

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

Correct answer is (c) $\frac{1}{12} n (n + 1)^{2} (n + 2)$

Q. The sum of n terms of 12+(12+22)+(12+22+32) +… is-

6n(n+1)(2n+1)​

6n(n+1)(2n−1)​

121​n(n+1)2(n+2)

121​n2(n+1)2

Correct answer is (c) 121​n(n+1)2(n+2)

Q. The sum of $n$ terms of $1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2})$ $+ \dots$ is-

$\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

$\frac{n ( n + 1 ) ( 2 n - 1 )}{6}$

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

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

Correct answer is (c) $\frac{1}{12} n (n + 1)^{2} (n + 2)$