Question

The coefficient of $x^{n-2}$ in the polynomial $(x-1)(x-2)(x-3)....(x-n)$ is

• A. $\frac{n(n^2+2)(3n+1)}{24}$
• B. $\frac{n(n^2-1)(3n+2)}{24}$
• C. $\frac{n(n^2+1)(3n+4)}{24}$
• D. None of these

Answer 1

The correct option is B $\frac{n(n^2-1)(3n+2)}{24}$

There are total of $n$ brackets. The term $x^{n-2}$ will be formed when integers are chosen from any two brackets and $x$ is chosen from all the other brackets and multiplied.

Thus, the Coefficient of $x^{n-2}$ is

$C=(1\times 2+1\times 3+...+1\times n)+(2\times 3+2\times 4+..+2\times n)+...+\left(\right(n-1)\times n)$

$=(\frac{\left(n\right)(n+1)}{2}-1)+2(\frac{\left(n\right)(n+1)}{2}-1-2)+...+(n-1)(\frac{\left(n\right)(n+1)}{2}-(1+2+3+..+(n-1)\left)\right)$

$=\left\{\right(1+2+3+...+(n-1)\left)\right(\frac{\left(n\right)(n+1)}{2}\left)\right\}-\{1+2(1+2)+3(1+2+3)+...+(n-1\left)\right(1+...+n-1\left)\right\}$

$=\left\{\right(\frac{(n-1)\left(n\right)}{2}\left)\right(\frac{\left(n\right)(n+1)}{2}\left)\right\}-\left\{{\sum }_1^{n-1}k\right(\frac{\left(k\right)(k+1)}{2}\left)\right\}$

$=\left\{\frac{n^2(n^2-1)}{4}\right\}-\left\{{\sum }_1^{n-1}\frac{k^3+k^2}{2}\right\}$

$=\left\{\frac{n^2(n^2-1)}{4}\right\}-\frac{1}{2}\left\{\right(\frac{(n-1)\left(n\right)}{2}{)}^2+\frac{(n-1)n(2n-1)}{6}\}$

$=\frac{n(n-1)}{4}\{n(n+1)-\frac{n(n-1)}{2}-\frac{2n-1}{3}\}$

$=\frac{n(n-1)}{4}\{\frac{n(n+3)}{2}-\frac{2n-1}{3}\}$

$=\frac{n(n-1)}{4}\left\{\frac{3n^2+9n-4n+2}{6}\right\}$

$=\frac{n(n-1)}{4}\left\{\frac{(3n+2)(n+1)}{6}\right\}$

$\therefore$ Option $B$ is correct.