1. induction question

For each m >=1, prove that

$
\displaystyle\sum_{k = 0}^{m} k . \dbinom{m}{k} = m \cdot 2 ^{m-1}$

No idea what to do

For each m >=1, prove that

$
\displaystyle\sum_{k = 0}^{m} k . \dbinom{m}{k} = m \cdot 2 ^{m-1}$

No idea what to do
$\displaystyle \sum_{k=0}^{m} k\cdot \dbinom{m}{k} = m \cdot 2 ^{m-1}$

proof:-
$\displaystyle \sum_{k=0}^{m} k\cdot \dbinom{m}{k} = 0 + \dbinom{m}{1} + 2 \dbinom{m}{2} + 3\dbinom {m}{3} + ... + m \dbinom{m}{m}$

$\displaystyle \sum_{k=0}^{m} k\cdot \dbinom{m}{k} = \frac{m!}{(m-1)!1!}+ \frac{2m!}{(m-2)!2!} + \frac{3m!}{(m-3)!3!} + ... + \frac{m!}{m!0!}$

$\displaystyle \sum_{k=0}^{m} k\cdot \dbinom{m}{k} = m \left( \frac{(m-1)!}{(m-1)!} + \frac{(m-1)!}{(m-2)!} + \frac{(m-1)!}{(m-3)!2!}+ ... +\frac{(m-1)!}{(m-1)!}\right) = m \cdot 2^{m-1}$

since
$\displaystyle 2^{m-1} = \sum_{k=0}^{m-1} \dbinom{m-1}{k}$

check here

ohh sorry by induction

let $P(m) : \sum_{k=0}^{m} k \dbinom{m}{k} = m2^{m-1}$

check for P(1)

$\sum_{k=0}^{1} k \dbinom{1}{k} = 0 + 1 \dbinom{1}{1}$ and this $= 1 \cdot 2^{0}$
P(m) is true for 1

suppose P(m) is true want P(m+1) true
want $\displaystyle \sum_{k=0}^{m+1} k \dbinom{m+1}{k} = (m+1)2^{m}$

$\displaystyle \sum_{k=0}^{m+1} k\cdot \dbinom{m+1}{k} = \frac{(m+1)!}{(m+1-1)!1!} +2 \frac{(m+1)!}{(m+1-2)!2!} +3 \frac{(m+1)!}{(m-2)!3!}+...+ \frac{(m+1)!}{(m+1)!}$

$\displaystyle \sum_{k=0}^{m+1} k\cdot \dbinom{m+1}{k} = (m+1)\left( \frac{(m)!}{(m+1-1)!1!} +2 \frac{m!}{(m+1-2)!2!} +3 \frac{(m)!}{(m-2)!3!}+...+ \frac{(m)!}{(m)!}\right)$

and $\displaystyle \frac{(m)!}{(m+1-1)!1!} +2 \frac{m!}{(m+1-2)!2!} +3 \frac{(m)!}{(m-2)!3!}+...+ \frac{(m)!}{(m)!} = \sum_{k=0}^{m} \dbinom{m}{k}$

$\displaystyle \sum_{k=0}^{m+1} k\cdot \dbinom{m+1}{k} = (m+1)\left(\sum_{k=0}^{m} \dbinom{m}{k}\right)= (m+1)2^m$

since $\displaystyle 2^m = \sum_{k=0}^{m} \dbinom{m}{k}$
so P(m) is true the proof ends

3. you are a GOD!
thanks so much Amer