Question

If $n$ is not a multiple of $3$, and ${\sum }_{r=0}^n{(-1)}^r{a}_r\left({{}^{n}C}_r\right)=k\left({{}^{n}C}_{\left[\frac{n}{3}\right]}\right)$, where $\left[x\right]$ denotes the greatest integer $\le x$, then $k$ equals

• A. $1$
• B. $0$
• C. $3$
• D. $-1$

Answer 1

The correct option is B $0$

${\sum }_{r=0}^{2n}{(-1)}^r\cdot a_r\cdot {{}^{n}C}_r$
$=$ coefficient of the constant term in
$[a_0+a_{1}x+....+a_{2n}{x}^{2n}]\times [{{}^{n}C}_0-{{}^{n}C}_1\left(\frac{1}{x}\right)+{{}^{n}C}_2{\left(\frac{1}{x}\right)}^2+.....+{(-1)}^n{{}^{n}C}_n{\left(\frac{1}{x}\right)}^n]$
$=$ coefficient of the constant term in
${(x^2+x+1)}^n{(1-\frac{1}{x})}^n$
$=$ coefficient of $x^n$ in ${(x^2+x+1)}^n{(x-1)}^n$
$=$ coefficient of $x^n$ in ${(x^3+1)}^n$
$=0$ as $n$ is not a multiple of $3$.