Given-n-N-prove-that-k-1-n-k-n-1-k-n-2-3-

Question Number 31713 by gunawan last updated on 13/Mar/18

Given n \in N

prove that

\underset{k = 1}{\sum^{n}} k (n + 1 - k) = (\begin{matrix} n + 2 \\ 3 \end{matrix})

Commented by abdo imad last updated on 13/Mar/18

w e h a v e \sum_{k = 1}^{n} k (n + 1 - k) = (n + 1) \sum_{k = 1}^{n} k - \sum_{k = 1}^{n} k^{2}

= (n + 1) \frac{n (n + 1)}{2} - \frac{n (n + 1) (2 n + 1)}{6}

= \frac{n (n + 1)}{2} (n + 1 - \frac{2 n + 1}{3}) = \frac{n (n + 1)}{2} (\frac{n + 2}{3})

= \frac{n (n + 1) (n + 2)}{6} f r o m a n o t h e r s i d e

C_{n + 2}^{3} = \frac{(n + 2)!}{3! (n + 2 - 3)!} = \frac{(n + 2)!}{6 {(n - 1)}^{'}} = \frac{(n + 2) (n + 1) n (n - 1)!}{6 (n - 1)!}

= \frac{n (n + 1) (n + 2)}{6} \Rightarrow \sum_{k = 1}^{n} k (n + 1 - k) = C_{n + 2}^{3} .

Commented by Tinkutara last updated on 13/Mar/18

I n o w u n d e r s t a n d y o u r n o t a t i o n!

Y o u w r i t e C_{n + 2}^{3} i n s t e a d o f^{n + 2} C_{3} .

B u t a c t u a l l y C_{n + 2}^{3} = 0 f o r n > 1 .

Commented by abdo imad last updated on 13/Mar/18

i h a v e u s e d t h e n o t a t i o n C_{n}^{p} = \frac{n!}{p! (n - p)!} f o r p ⩽ n .