Statement-1: The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400) is 8000Statement-2: ∑k=1n k3−(k−1)3=n3 for each natural number n.

A
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
B
STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
C
STATEMENT-1 is True, STATEMENT-2 is False
D
STATEMENT-1 is False, STATEMENT-2 is True

Solution:

Statement $- 2$ is true since
$\begin{array}{l} \sum_{k = 1}^{n} (k^{2} - (k - 1)^{3}) = (1^{3} - 0^{3}) + (2^{3} - 1^{3}) + \dots \\ + (n^{3} - (n - 1)^{3}) = n^{3} \\ \Rightarrow \sum_{k = 1}^{n} (k - (k - 1)) ((k^{2} + k (k - 1) + (k - 1)^{2}) = n^{3} \\ \Rightarrow \sum_{k = 1}^{n} (k^{2} + k (k - 1) + (k - 1)^{2}) = n^{3} \end{array}$

Putting $n = 20,$ we get

$\sum_{k = 1}^{20} ((k - 1)^{2} + k (k - 1) + k^{2}) = 20^{3}$
$\begin{array}{l} \Rightarrow 1 + (1 + 2 + 4) + (4 + 6 + 9) + \dots + \\ (361 + 380 + 400) = 8000 \end{array}$

Thus, Statement $- 1$ is also true and Statement $- 2$ is a correct reason for it.

Statement $- 1$ : The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$
Statement $- 2$ : $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ for each natural number $n .$

Solution:

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