Question

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

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

Answer 1

Answer: (B)

Let, $E=\left(x-{\left(\alpha \right)}_1\right)\left(x-{\left(\alpha \right)}_2\right)\left(x-{\left(\alpha \right)}_3\right)\dots \left(x-{\left(\alpha \right)}_n\right)$ where ${\alpha }_1=1,{\alpha }_2=2$ etc.
$=x^n-\left(\sum {\left(\alpha \right)}_1\right)x^{n-1}+\left(\sum {\left(\alpha \right)}_1{\left(\alpha \right)}_2\right)x^{n-2}+...$
Hence, the coefficient of $x^{n-2}=$ the sum of all the products of the first $‘n’$ natural numbers taken two at a time
$\frac{{\left(1+2+3+.\dots \dots +n\right)}^2-\left(1^2+2^2+3^2+.\dots \dots .+n^2\right)}{2}$ $=\frac{n\left(n^2-1\right)\left(3n+2\right)}{24}$