prove-that-1-x-n-1-nx-n-n-1-2-x-2-n-n-1-n-2-3-x-3-n-n-n-using-a-suitable-expansion-method-hence-determine-the-expansion-of-2-001-89-

Question Number 55815 by Rio Mike last updated on 04/Mar/19

p r o v e t h a t

{(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots n (n - n)

u s i n g a s u i t a b l e e x p a n s i o n m e t h o d

h e n c e d e t e r m i n e t h e e x p a n s i o n o f

{(2.001)}^{89}

Answered by Kunal12588 last updated on 04/Mar/19

w e c a n e x p a n d {(1 + x)}^{n} u s i n g b i n o m i a l

e x p a n s i o n .

{(a + b)}^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} b^{1} + \dots +^{n} C_{r} a^{n - r} b^{r} + \dots +^{n} C_{n} b^{n}

p u t a = 1 a n d b = x

{(1 + x)}^{n} =^{n} C_{0} 1^{n} +^{n} C_{1} 1^{n - 1} x + \dots +^{n} C_{n} x^{n}

= 1 + \frac{n (n - 1)!}{(n - 1)!} x + \frac{n (n - 1) (n - 2)!}{2! (n - 2)!} x^{2} + \dots + x^{n}

= 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots + x^{n}

{(2.001)}^{89} = {(1 + 1.001)}^{89}

= 1 + 89 (1.001) + \frac{89 \times 88}{2} {(1.001)}^{2} + \dots + {(1.001)}^{89}