Question

If $A{=}^{2n}{C}_0{.}^{2n}{C}_1{+}^{2n}{C_1}^{2n-1}{C}_1{+}^{2n}{C_2}^{2n-2}{C}_1+...$ then $A$ is

• A. $0$
• B. $n{.2}^{2n}$
• C. $2^{10}-2$
• D. $1$

Answer 1

Answer: (B)

$n{.2}^{2n}$

Given $A{=}^{2n}{C}_0{.}^{2n}{C}_1{+}^{2n}{C}_1{.}^{2n-1}{C}_1{+}^{2n}{C}_2{.}^{2n-2}{C}_1+........$

we can write $A$ as summation .

$\Rightarrow A={\sum }_{r=0}^{2n}{.}^{2n}{C}_r{.}^{2n-r}{C}_1={\sum }_{r=0}^{2n}.\frac{\left(2n\right)!}{r!(2n-r)!}.\frac{(2n-r)!}{(2n-r-1)!1!}={\sum }_{r=0}^{2n}\frac{\left(2n\right)!}{r!(2n-r-1)!}=2n{\sum }_{r=0}^{2n}{}^{}{.}^{2n-1}{C}_r$

$=2n{.2}^{2n-1}=n{.2}^{2n}$.

$\therefore A=n{.2}^{2n}$