in the binomial expansion of (1+x/k)^n, where k is a constant and n is a positive integer, the coefficients of x and x^2 are equal.

Question

(a) show that 2k=n-1
(b) deduce the value of k. Hence find the 1st 3 terms in the expansion in ascending powers of x.

Answer 1

(1+x/k)^n = 1^n + n(x/k) + n(n-1)/2 (x/k)^2
n/k = n(n-1)/2k^2
2kn/2k^2 = n(n-1)/2k^2
2kn = n(n-1)
2k = n-1

we know that n is odd, since n=2k+1
if k=1,
(1+x)^3 = 1+3x+3x^2+...

if k=2, we have
(1+x/2)^5 = 1+5(x/2) + 5*4/2*(x/2)^2+...
= 1 + 5/2 x + 5/2 x^2 + ...

For any integer value of k>0, n=2k+1
will work.

So, (b) is really a bogus question