Question-178500

Question Number 178500 by infinityaction last updated on 17/Oct/22

Commented by abdullahoudou last updated on 17/Oct/22

i t w o u l d b e n i c e t o g i v e t h e r e s t o f q u e s t i o n s i n s t e a d o f o n l y n u m b e r 15 \dots

Answered by mindispower last updated on 17/Oct/22

== \sum_{k ⩾ 0} \frac{k (k - 1) n!}{(n - k)! k!}

= \underset{k = 0}{\sum^{n}} k (k - 1) C_{k}^{n}

\underset{k = 0}{\sum^{n}} x^{k} C_{n}^{k} = {(1 + x)}^{n}

\Rightarrow \sum_{k ⩾ 1} {k x}^{k - 1} C_{k}^{n} = n {(1 + x)}^{n - 1}

Σ k (k - 1) x^{k - 2} = n (n - 1) {(1 + x)}^{n - 2}

Σ k (k - 1) C_{n}^{k} = Σ \frac{n!}{(n - k)! (k - 2)!} = n (n - 1) 2^{n - 2}

Commented by infinityaction last updated on 17/Oct/22

t h a n k y o u s i r

Commented by mindispower last updated on 02/Nov/22

w i t h e p l e a s u r