Question

If the value of $C_0+2\cdot C_1+3\cdot C_2+\cdots +(n+1)\cdot C_n=576$ ,then n is equal to

• A. 7
• B. 5
• C. 6
• D. 9

Answer 1

Answer: (A)

Given, $C_0+2C_1+3C_2+\dots +(n+1)C_n=576$
We know that,
$(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+\dots +{}^n{C}_n{x}^n$
$\Rightarrow x(1+x{)}^n={}^n{C}_{0}x+{}^n{C}_1{x}^2+{}^n{C}_2{x}^3+\dots +{}^n{C}_n{x}^{n+1}$
On differentiating w.r.t. $x$, we get
$(1+x{)}^n+x\cdot n(1+x{)}^{n-1}$
$={}^n{C}_0+2\cdot {}^n{C}_1\cdot x+3{}^n{C}_2{x}^2+\dots +(n+1){}^n{C}_n{x}^n$
On putting $n=1$, we get
$2^n+n\cdot 2^{n-1}={}^n{C}_0+2\cdot {}^n{C}_1+3\cdot {}^n{C}_1+\dots +(n+1){}^n{C}_n$
$\Rightarrow 2^{n-1}(n+2)=576$ (given)
$\Rightarrow 2^{n-1}(n+2)=2^6\times 9=2^{(7-1)}\cdot (7+2)$
On comparing, we get $n=7$