Question

Find the value of ${}^n{C}_1+2^n{C}_2+3^n{C}_3.........n^n{C}_n$

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

Answer 1

The correct option is D

$n\cdot 2^{n-1}$

Method 1(By differentiation)
${\left(1+x\right)}^n=nC0+nC1x+nC2x^2+....+nCn{x}^n$
Differentiate both the sides
$n{\left(1+x\right)}^{n-1}=nC1+2nC2x+3nC3x^2+....+nnCn{x}^{n-1}$

⇒ put x = 1

$n{2}^{n-1}=nC1+2nC2+3nC3+....+nnCn$

Method 2

Each term of the sum is r ${}^n{C}_r$, where r varies from 1 to n

${}^n{C}_r=\frac{n}{r}$ ${}^{n-1}{C}_{r-1}$

⇒ $r^n{C}_r=n^{n-1}{C}_{r-1}$

$nC1+2nC2+3nC3+....+nnCn$$=\sum _{r=1}^{n}r{\cdot }^n{C}_r$

= $\sum _{r=1}^{n}n{\cdot }^{(n-1)}{C}_{(r-1)}$

$=n\cdot \sum _{r=1}^n$ ${}^{(n-1)}{C}_{(r-1)}$

= $n\cdot {[}^{n-1}{C}_0$ + ${}^{n-1}{C}_1$..................... ${}^{n-1}{C}_{n-1}]$

= $n{2}^{n-1}$

[Because $(1+x{)}^{n-1}$ = ${}^{n-1}{C}_0$ + ${}^{n-1}{C}_{1}x$ ..........${}^{n-1}{C}_{n-1}$$x^{n-1}$
If we put $x=1$, we get
$\Rightarrow 2^{n-1}={}^{n-1}{C}_0$ + ${}^{n-1}{C}_1$ + .............${}^{n-1}{C}_{n-1}$