Question

If $\frac{{}^n{C}_0}{2^n}+2.\frac{{}^n{C}_1}{2^n}+3.\frac{{}^n{C}_2}{2^n}+\dots .(n+1)\frac{{}^n{C}_n}{2^n}=16$, then the value of '$n$' is:

• A. $20$
• B. $25$
• C. $30$
• D. $40$

Answer 1

The correct option is C $30$

$(1+x{)}^n=1+x{}^n{C}_1+x^2{}^n{C}_2...+x^n{}^n{C}_n$
Multiplying with $x$ on both the sides, we get
$x(1+x{)}^n=x+x^2{}^n{C}_1+x^3{}^n{C}_2...+x^{n+1}{}^n{C}_n$
Now differentiating with respect to $x$ on both the sides, we get.
$nx(1+x{)}^{n-1}+(1+x{)}^n=1+2x{}^n{C}_1+3x^2{}^n{C}_2...+(n+1)x^n{}^n{C}_n$
$(1+x{)}^{n-1}(1+x+nx)=1+2x{}^n{C}_1+3x^2{}^n{C}_2...+(n+1)x^n{}^n{C}_n$
Substituting $x=1$ in (i), we get.
$2^{n-1}(n+2)=1+2{}^n{C}_1+3{}^n{C}_2...+(n+1){}^n{C}_n$
Dividing both the side by $2^n$, we get.
$\frac{2^{n-1}(n+2)}{2^n}=\frac{1}{2^n}[1+2{}^n{C}_1+3{}^n{C}_2...+(n+1\left){}^n{C}_n\right]$

$\frac{n+2}{2}=16$ ... from the given question
$n+2=32$
$n=30$