Binomial expansion

Question

The first three terms in ascending powers of x, in the expansion of (1 + a/x)ⁿ are 1 + 24/x + 252/x³.

How do I find the values of a and n?

Answer 1

(1 + a/x)ⁿ = (_nC₀)(1)ⁿ(a/x)⁰ + (_nC₁)(1)^n-1(a/x)¹ + (_nC₂)(1)^n-2(a/x)² + ... + (_nC_n)(1)⁰(a/x)ⁿ

= 1 + 24/x + 252/x² + ...

_nC₀ = n! / ((0!)(n-0)!) = 1

_nC₁ = n! / ((1!)(n-1)!) = n So, n(a/x) = 24/x. Therefore, an = 24 (equivalently, n = 24/a)

_nC₂ = n! / ((2!)(n-2)!) = n(n-1)/2 So, n(n-1)/2 = 252 / a²

(1/2)(24/a)(24/a - 1) = 252/a²

So, 12(24-a) = 252.

24 - a = 21

a = 3 and n = 24/a = 8

Answer 2

a=3 and n=8, but the third term is not x³, it's x².

Answer 3

(1+a/x)^n = 1 + nC1(a/x) + nC2(a/x)^2 + ...

na=24

nC1(a) = 24

nC1= n =24/a, a=24/n

nC2(a^2) =252= n(n-1)/2)(a^2)

(24/a)(24/a- 1)(1/2)(a^2) = 252

12a(24/a - 1)= 252

24-a = 21

a =3, n=8

(1+ 3/x)^8 = 1 + 8(3/x) +28(3/x)^2 + ...

= 1+24/x + 28(9)/x^2 + ...

= 1+24/x +252/x^2 + ....