If-1-2x-x-2-n-r-0-2n-a-r-x-r-then-a-r-

Question Number 56137 by gunawan last updated on 11/Mar/19

If {(1 + 2 x + x^{2})}^{n} = \underset{r = 0}{\sum^{2 n}} a_{r} x^{r}, then a_{r} =

Commented by maxmathsup by imad last updated on 11/Mar/19

{(1 + 2 x + x^{2})}^{n} = {(x + 1)}^{2 n} = \sum_{r = 0}^{2 n} C_{2 n}^{r} x^{r} \Rightarrow a_{r} = C_{2 n}^{r} = \frac{(2 n)!}{r! (2 n - r)!}

a n d r \in [[0, 2 n]] .

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19

{(1 + 2 x + x^{2})}^{n}

{(1 + x)}^{2 n} = a_{0} x^{0} + a_{1} x^{1} + a_{2} x^{2} + \dots + a_{2 n} x^{2 n}

p u t x = 1

2^{2 n} = \underset{r = 0}{\sum^{2 n}} a_{r}

r + 1 t h e t e r m 2 {n c}_{r} x^{2 n}

s o a_{r} = 2 {n c}_{r} = \frac{2 n!}{r! (2 n - r)!}