Question

If the coefficients of xr-1 , xr , xr+1 in the binomial expansion of (1+x)n are in A.P. , prove that

n2 - n(4r + 1) + 4r2 - 2 = 0 .

Answer 1

Answer:

Step-by-step explanation:

Hi,

Let Coefficient of x^(r-1) in the expansion of (1 + x)ⁿ be a = ⁿCr-1

Coefficient of x^(r) in the expansion of (1 + x)ⁿ be b = ⁿCr

Coefficient of x^(r+1) in the expansion of (1 + x)ⁿ be c = ⁿCr+1

Given that a, b and c are in A.P,

so 2b = a + c

2ⁿCr = ⁿCr-1 + ⁿCr+1

2(n!/(n-r)!r!) = n!/(n-r+1)!(r-1)! + n!/(n-r-1)!(r+1)!

Multiplying the equation by  (n-r + 1)!(r+1)!, and dividing by n! we get

2(n- r + 1)(r + 1) = r(r + 1) + (n - r + 1)(n - r)

On simplifying , we get

2[nr + n - r² - r + r + 1] = r² + r + n² - nr - nr + r² + n - r

On simplifying and rearranging terms , we get

4r² + n² - 4rn - n - 2 = 0

or

n² - n(4r + 1) + 4r² - 2 = 0

Hence, Proved

Hope, it helps !