In the binomial expansion of (a-b)^n, n ≥5, the sum of 5th and 6th terms is 0, then find a/b

Question

binomial theorem (cbse

Answer 1

expand (a - b)^n. We get,

a^n*(1 + (-b/a))^n = a^n{1 + n(-b/a) + n(n-1)(-b/a)^2/2 + n(n-1)(n-2)(-b /a)^3/3! + n(n-1)(n-2)(n-3)(-b /a)^4/4! + n(n-1)(n-2)(n-3)(n-4)(-b /a)^5/5! + ... }

The sum of the 5th and 6th terns is zero, so then

n(n-1)(n-2)(n-3)(-b /a)^4/4! + n(n-1)(n-2)(n-3)(n-4)(-b /a)^5/5! = 0

Cancelling out common factors,

(-b /a)^4/4! + (n-4)(-b /a)^5/5! = 0

Cancelling out more common factors,

1/4! + (n-4)(-b /a)/5! = 0

1 + (n-4)(-b/a)/5 = 0

5 = (n-4)(b/a)

(a/b) = (n-4)/5, n ≥5