expanding the binomial ... (1 + px)^n = nC0 1^n + nC1 1^(n-1) (px)^1 +
nC2 1^(n-2) (px)^2 + ...
2nd term ... n p = 15
3rd term ... [n (n - 1) / 2] p^2 = 90
dividing terms ... (n - 1) p = 12 ... n p - p = 12
subtracting equations ... p = 3
substitute back to find n
If (1+px)^n = 1 + 15x +90x^2 +... , find the values of p and n.
Answer: p=3, n=5
Thanks for any help
1 answer
1 answer