Question

If $S={\sum }_{n=1}^{\infty }\left(\frac{{}^n{C}_0{+}^n{C}_1{+}^n{C}_2+....{+}^n{C}_n}{{}^n{P}_n}\right),$ then $S=$

• A. $2e$
• B. $2e-1$
• C. $2e+1$
• D. None of these

Answer 1

The correct option is D None of these

${\sum }_{n=1}^{\infty }\frac{{}^n{C}_0{+}^n{C}_1+..{.}^n{C}_n}{{}^n{P}_n}$
$=\frac{{}^1{C}_0{+}^1{C}_1}{{}^1{P}_1}+\frac{{}^2{C}_0{+}^2{C}_1{+}^2{C}_2}{{}^2{P}_2}+...$
$=\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}+...=(1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!}...)-1=e^2-1$