Which of the following statements is/are correct?

​$$\text {(a) } \Sigma\left(n^2-n\right)=330, \mathrm{n}=1,2, \ldots, 10$$​​

​$$\text {(b) } \Sigma\left(n^2+n\right)=728, \mathrm{n}=1,2, \ldots, 12$$​​

​$$\text {(c) } \Sigma\left(n^2+3 n+1\right)=352, \mathrm{n}=1,2, \ldots, 8$$

Correct option is A

Given:

​$$\text {(a) } \Sigma\left(n^2-n\right)=330, \mathrm{n}=1,2, \ldots, 10$$​​

​$$\text {(b) } \Sigma\left(n^2+n\right)=728, \mathrm{n}=1,2, \ldots, 12$$​​

​$$\text {(c) } \Sigma\left(n^2+3 n+1\right)=352, \mathrm{n}=1,2, \ldots, 8$$

Formula:

$$\textbf{Standard Summation Formulas:} \\\sum n = \frac{n(n+1)}{2},\\\ \\ \quad \sum n^2 = \frac{n(n+1)(2n+1)}{6}$$​​

Solution:

$$\textbf{(a) } \sum_{n=1}^{10} (n^2 - n) = \sum n^2 - \sum n \\\sum n^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \\\ \\\sum n = \frac{10 \cdot 11}{2} = 55 \\\ \\\Rightarrow \sum (n^2 - n) = 385 - 55 = 330 \quad \text{ Correct}\\[10pt]\textbf{(b) } \sum_{n=1}^{12} (n^2 + n) = \sum n^2 + \sum n \\\ \\\sum n^2 = \frac{12 \cdot 13 \cdot 25}{6} = 650 \\\ \\\sum n = \frac{12 \cdot 13}{2} = 78 \\\ \\\Rightarrow \sum (n^2 + n) = 650 + 78 = 728 \quad \text{ Correct}\\[10pt]\textbf{(c) } \sum_{n=1}^{8} (n^2 + 3n + 1) = \sum n^2 + 3\sum n + \sum 1 \\\ \\\sum n^2 = \frac{8 \cdot 9 \cdot 17}{6} = 204 \\\ \\\sum n = \frac{8 \cdot 9}{2} = 36 \\\ \\\sum 1 = 8 \\\Rightarrow \sum (n^2 + 3n + 1) = 204 + 3 \cdot 36 + 8 = 204 + 108 + 8 = 320 \\\text{Given: 352} \Rightarrow \text{ Incorrect}$$

​​Correct Answer: Option A (a) and (b)