Question

If $C_r={}^{2n+1}C_r$ then, $C_0-{C_1}^2+{C_2}^2+....+(-1){{}^{2n+1}C^2}_{2n+1}$ is equal to?

• A. ${\left({}^{2n+1}C_n-{}^{2n+1}C_{2n+1}\right)}^2$
• B. ${}^{2n}C_n$
• C. $\frac{1}{n}\left({}^{2n}C_n\right)$
• D. $0$

Answer 1

The correct option is D

$0$

Find the value of ${\mathit{C}}_{\mathbf{0}}\mathbf{}\mathbf{-}\mathbf{}{{\mathbf{C}}_{\mathbf{1}}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{{\mathbf{C}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}{{}^{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{C}^{\mathbf{2}}}_{\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}}$:

Given, $C_r={}^{2n+1}C_r$

As we know,

${\mathbf{(}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{)}}^{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{C}}_{\mathbf{0}}\mathbf{}\mathbf{-}\mathbf{}{\mathit{C}}_{\mathbf{1}}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}{\mathit{C}}_{\mathbf{2}}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{}{\mathit{C}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}$

Put $x=1$

$0=C_0-C_1+C_2-.....+C_{\mathrm{n}}$

Now,

${(1-\mathrm{x})}^{\mathrm{n}}{(1-\mathrm{x})}^{\mathrm{n}}=\left(C_0-C_{1}x+C_2{x}^2-.....+C_{\mathrm{n}}{x}^{\mathrm{n}}\right)\left(C_0-C_{1}x+C_2{x}^2-.....+C_{\mathrm{n}}{x}^{\mathrm{n}}\right)$

${\left(1-x\right)}^{2n}=\left(C_0-{C_1}^{2}x+{C_2}^2{x}^2-.....+{C_{\mathrm{n}}}^2{x}^{\mathrm{n}}+{\left(-1\right)}^{2n+1}{C^2}_{2n+1}\right)$

Again put $x=1$

$\left(C_0-{C_1}^2+{C_2}^2-.....+{C_n}^2+{\left(-1\right)}^{2n+1}{C^2}_{2n+1}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{0}$

Hence, Option ‘D’ is Correct.