Question

Value of $C_0+2C_1+3C_2+4C_3+\dots +(n+1)C_n$ is
( 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 $(n+2)\cdot 2^{n-1}$

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

Alternate solution:

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\dots +C_n{x}^n$
$x(1+x{)}^n=C_{0}x+C_1{x}^2+C_2{x}^3+C_3{x}^4+\dots +C_n{x}^{n+1}$

Differentiating w.r.t. $x$ on both sides, we get
$xn(1+x{)}^{n-1}+(1+x{)}^n=C_0+2C_{1}x+3C_2{x}^2+4C_3{x}^3+\dots +(n+1)C_n{x}^n$
By putting $x=1,$ we get
$C_0+2C_1+3C_2+4C_3+\dots +(n+1)C_n=n\cdot 2^{n-1}+2^n=(n+2)\cdot 2^{n-1}$