Question

The coefficient of $x^{n - 2}$ in the polynomial $\left(x - 1\right)\left(x - 2\right)\left(x - 3\right)\ldots \ldots \ldots \left(x - n\right)$ is

• A. $\frac{n \left(n^{2} + 2\right) \left(3 n + 1\right)}{24}$
• B. $\frac{n \left(n^{2} - 1\right) \left(3 n + 2\right)}{24}$
• C. $\frac{n \left(n^{2} + 1\right) \left(3 n + 4\right)}{24}$
• D. $\frac{n \left(n^{2} - 2\right) \left(3 n - 2\right)}{24}$

Answer 1

Answer: (B)

Let, $E=\left(x - \left(\alpha \right)_{1}\right)\left(x - \left(\alpha \right)_{2}\right)\left(x - \left(\alpha \right)_{3}\right)\ldots \left(x - \left(\alpha \right)_{n}\right)$ where $\alpha _{1}=1,\alpha _{2}=2$ etc.
$=x^{n}-\left(\displaystyle \sum \left(\alpha \right)_{1}\right)x^{n - 1}+\left(\displaystyle \sum \left(\alpha \right)_{1} \left(\alpha \right)_{2}\right)x^{n - 2}+...$
Hence, the coefficient of $x^{n - 2}=$ the sum of all the products of the first $‘n’$ natural numbers taken two at a time
$\frac{\left(1 + 2 + 3 + . \ldots \ldots + n\right)^{2} - \left(1^{2} + 2^{2} + 3^{2} + . \ldots \ldots . + n^{2}\right)}{2}$ $=\frac{n \left(n^{2} - 1\right) \left(3 n + 2\right)}{24}$