Question

If $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+....C_n{x}^n,$ then ${\sum }_{r=0}^n{\sum }_{s=0}^n(r+s)C_r{C}_s$ is equal to :

• A. $2^{2n}$
• B. $n{.2}^{2n}$
• C. $n{.2}^n$
• D. None of these

Answer 1

The correct option is B $n{.2}^{2n}$

$I=\sum _{r=0}^n\sum _{s=0}^n(r+s)C_r{C}_s$

$=\sum _{r=0}^n\sum _{s=0}^n{}^r{C}_r{C}_s+\sum _{r=0}^n\sum _{s=0}^n{C}_r{C}_s$

$I=\sum _{r=0}^n{}^r{C}_r.\sum _{s=0}^n{C}_3{+}^n\sum _{s=0}S{C}_s.\sum _{r=0}^n{C}_r$

(A)

$A=\sum _{r=0}^n{}^r{C}_r.\sum _{s=0}^n{c}^s$

$(1+x{)}^n=C_0+C_{1}x+...$

Put $x+1$

$2^n=\sum _{r=0}^n{C}_s$...(1)

$(1+x{)}^n=C_0+C_1+x.....C_n{x}^n$

differentiate with respect to $x$

$n(1+x{)}^{n-1}=C_1+2C_{2}x+3C_3{x}^2+....$

Put $x=1$

$n(2{)}^{n-1}=\sum _{r=0}^n{}^{r}Cr$

(B)

$A=\sum _{r=0}^n{}^r{C}_r.\sum _{s=0}^n{C}_5$

$A=n{2}^{n-1}{.2}^n$ from(1) and (2)

$B=\sum _{s=0}^n{}^s{C}_n.\sum _{r=0}^n{C}_r$

$B=n{2}^{n-1}{.2}^n$ from(1) and (2)

$I=(A+B-n{.2}^n{.2}^{n-1}).2$

$I=n{.2}^{2n}$

$\therefore I+n{.2}^{2n}$