let-j-e-i-2pi-3-find-the-value-of-k-0-n-C-n-k-1-j-k-j-2n-2k-

Question Number 30519 by abdo imad last updated on 22/Feb/18

l e t j = e^{i \frac{2 π}{3}} f i n d t h e v a l u e o f \sum_{k = 0}^{n} C_{n}^{k} {(1 + j)}^{k} j^{2 n - 2 k} .

Answered by sma3l2996 last updated on 23/Feb/18

\underset{k = 0}{\sum^{n}} C_{n}^{k} {(1 + j)}^{k} j^{2 n - 2 k} = \underset{k = 0}{\sum^{n}} C_{n}^{k} {(1 + j)}^{k} j^{2 (n - k)}

= \underset{k = 0}{\sum^{n}} C_{n}^{k} {(1 + j)}^{k} {(j^{2})}^{n - k} = {(1 + j + j^{2})}^{n} = 0 b e c a u s e (j^{2} + j + 1 = 0)