Question

Find the sum $2C_o+\frac{2^2}{2}{C}_1+\frac{2^3}{3}{C}_2+\frac{2^4}{4}{C}_3+...+\frac{2^{11}}{11}{C}_{10}$

• A. $\frac{3^{11}}{11}$
• B. $\frac{3^{11}-1}{11}$
• C. $\frac{3^{10}-1}{11}$
• D. $\frac{3^{10}+1}{11}$

Answer 1

Answer: (B)

$\frac{3^{11}-1}{11}$

Consider $(1+x{)}^{10}$
$(1+x{)}^{10}=1+{}^{10}{C}_1{x}^1+{}^{10}{C}_2{x}^2...+{}^{10}{C}_{10}{x}^{10}$
Integrating both sides with respect to x, we get
$\frac{(1+x{)}^{11}}{11}+C=x+\frac{{}^{10}{C}_1{x}^2}{2}+\frac{{}^{10}{C}_2{x}^3}{3}...\frac{{}^{10}{C}_{10}{x}^{11}}{11}$
For the value of C, we substitute $x=0$
Hence we get
$\frac{1}{11}+C=0$
$C=-\frac{1}{11}$
Therefore the entire equation re-frames as
$\frac{(1+x{)}^{11}-1}{11}=x+\frac{{}^{10}{C}_1{x}^2}{2}+\frac{{}^{10}{C}_2{x}^3}{3}...\frac{{}^{10}{C}_{10}{x}^{11}}{11}$
Now substituting $x=2$ in the above equation, we get
$2+\frac{{}^{10}{C}_1{2}^2}{2}+\frac{{}^{10}{C}_2{2}^3}{3}...+\frac{{}^{10}{C}_{10}{2}^{11}}{11}=\frac{(1+2{)}^{11}-1}{11}$
$=\frac{3^{11}-1}{11}$