Question

${\sum }_{k=0}^{n}k\cdot {}^n{C}_k=n\cdot 2^{n-1},nϵN$

Answer 1

P(1) = ${\sum }_{k=0}^{m}k{}^1{C}_k=0+{1.}^1{C}_1={1.2}^0=1$
Hence the given statement holds for n = 1
Now assume that it holds for n = m
$\therefore P\left(1\right)={\sum }_{k=0}^{m}k\cdot {}^m{C}_k=m\cdot 2^{m-1}$ .....(1)
Now P(m + 1) = ${\sum }_{k=0}^{m+1}k\cdot {}^{m+1}{C}_k$
= ${\sum }_{k=0}^{m+1}k{[}^m{C}_{k-1}+{}^m{C}_k]$
= ${\sum }_{k=0}^{m+1}k{}^m{C}_{k-1}+{\sum }_{k=0}^{m+1}k\cdot {}^m{C}_k$
The first summation is meaning less for k = 0 as ${}^m{C}_{-1}$
has no meaning and the second is meaningless for k = m + 1 as ${}^m{C}_{m+1}$ is also meaningless.
$P(m+1)={\sum }_{k=1}^{m+1}k{}^m{C}_{k-1}+{\sum }_{k=0}^{m}k{}^m{C}_k$
Put k = p + 1 in 1st sigma and adjust the limits
$\therefore P(m+1)={\sum }_{p=0}^m(p+1){}^m{C}_p+{\sum }_{k=0}^{m}k{}^m{C}_k$
$={\sum }_{k=0}^m(k+1){}^m{C}_k+{\sum }_{k=0}^{m}k\cdot {}^m{C}_k$
$={\sum }_{k=0}^m(2k+1){}^m{C}_k$
$=2{\sum }_{k=0}^{m}k{}^m{C}_k+{\sum }_{k=0}^m{}^m{C}_k$
$=2(m\cdot 2^{m-1})+2^m=m\cdot 2^m+2^m$
$=(m+1)2^m$
$\therefore$The statement holds good for n = m + 1