Question

If $n=11$ then $C_0^2-C_1^2+C_2^2-C_3^2+....+{(-1)}^n{C}_n^2$ equals

• A. ${(-1)}^5{{}^{10}C}_5$
• B. $0$
• C. ${{}^{10}C}_6$
• D. None of these

Answer 1

The correct option is B $0$

$S={\sum }_0^n(-1{)}^r{C}_r^2$

where $n=11$

We know that:

$(1+x{)}^n={\sum }_0^n{C}_r{x}^r$

& $(1-x{)}^n={\sum }_0^n{C}_r(-x{)}^r$

So,

$(1+x{)}^n(1-x{)}^n=\left({\sum }_0^n{C}_r{x}^r\right)\left({\sum }_0^n{C}_r\right(-x{)}^r)$----------- $(1)$

The coefficient of $x^n$ in eq. $\left(1\right)$ is ${\sum }_0^n(-1{)}^r{C}_r^2$ which is equal to $S$

$(1+x{)}^n(1-x{)}^n=(1-x^2{)}^n={\sum }_0^n{C}_r(-x^2{)}^n$

$(1-x^2{)}^n={\sum }_0^n{C}_r(-1{)}^r(x{)}^{2r}$ ------------- $(2)$

Put $r=\frac{n}{2}$ in Eq. $\left(2\right)$:

Coefficient of $x^n$ is: $(-1{)}^{\frac{n}{2}}{C}_{\frac{n}{2}}^n$ (If $n$ is even)

0 (If $n$ is odd)

Since $n=11$,

$S={\sum }_0^n(-1{)}^r{C}_r^2=0$

Option $B$