Question

The value of $C_0+2C_1+3C_2+4C_3+\dots +(n+1)C_n$ is
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> ( where $C_r={}^n{C}_r$)

• A. $(n+2)\cdot 2^n$
• B. $n\cdot 2^n$
• C. $n\cdot 2^{n-1}$
• D. $(n+2)\cdot 2^{n-1}$

Answer 1

The correct option is D $(n+2)\cdot 2^{n-1}$

$S={}^n{C}_0\cdot a+{}^n{C}_1\cdot (a+d)+{}^n{C}_2\cdot (a+2d)+\cdots {+}^n{C}_n(a+nd)$
We know that
$S=\frac{a+(a+nd)}{2}\cdot 2^n$
Then
$C_0+2C_1+3C_2+4C_3+\dots +(n+1)C_n=\frac{1+n+1}{2}\cdot 2^n=(n+2)2^{n-1}$

Alternate solution
Let $S_n=C_0+2C_1+3C_2+4C_3+\dots +(n+1)C_n$
$=\sum _{r=0}^n(r+1)C_r=\sum _{r=0}^{n}r{C}_r+\sum _{r=0}^n{C}_r=\sum _{r=1}^{n}r{C}_r+\sum _{r=0}^n{C}_r=n\sum _{r=1}^n{}^{n-1}{C}_{r-1+}+\sum _{r=0}^n{}^n{C}_r=n\cdot 2^{n-1}+2^n=(n+2)\cdot 2^{n-1}$