Question

The expressions ${\left({}^{10}{C}_0\right)}^2-{\left({}^{10}{C}_1\right)}^2+{\left({}^{10}{C}_2\right)}^2-\dots \dots .+{\left({}^{10}{C}_8\right)}^2-{\left({}^{10}{C}_9\right)}^2+{\left({}^{10}{C}_{10}\right)}^2$ equals

• A. 10 !
• B. ${\left({}^{10}{C}_5\right)}^2$
• C. $-{}^{10}{C}_5$
• D. ${}^{10}{C}_5$

Answer 1

Answer: (C)

We have
$(1+x{)}^{10}={}^{10}{C}_0+{}^{10}{C}_{1}x+{}^{10}{C}_2{x}^2+{}^{10}{C}_3{x}^3+\dots \dots ..+{}^{10}{C}_9{x}^9+{}^{10}{C}_{10}{x}^{10}$....(1)
$\text{ Also }(x-1{)}^{10}={}^{10}{C}_0{x}^{10}-{}^{10}{C}_1{x}^9+{}^{10}{C}_2{x}^8+\dots \dots ..{}^{10}{C}_{9}x+{}^{10}{C}_{10}$....(2)
$\text{ Multiplying (1) and }(2)\text{, we get }$
${\left(x^2-1\right)}^{10}=\left({}^{10}{C}_0+{}^{10}{C}_{1}x+\dots \dots .+{}^{10}{C}_9{x}^9+{}^{10}{C}_{10}{x}^{10}\right)$
$\left({}^{10}{C}_0{x}^{10}+{}^{10}{C}_1{x}^9+\dots \dots .-{}^{10}{C}_{9}x+{}^{10}{C}_{10}\right)$
Comparing the coefficients of $x^{10}$ in (3), we get
${}^{10}{C}_5(-1{)}^5={\left({}^{10}{C}_0\right)}^2-{\left({}^{10}{C}_1\right)}^2+{\left({}^{10}{C}_2\right)}^2-\dots \dots .-{\left({}^{10}{C}_9\right)}^2+{\left({}^{10}{C}_{10}\right)}^2$