the first three terms in asscending powers of x of the binomial exapansion (1+bx)^n are 1 + 28x + 336x^2
n is a psotive interger find the value of n and b
good luck
if it is any help i got as far as 1. bx^n = 28x and 2. 672=(n^2-n)b^2
(1+bx)^n=1+nbx+n(n-1)/2.(bx)^2..... (1+bx)^n=1+(x)(nb)+(x^2)[0.5n(n-1)b^2]..... (1+bx)^n=1+28x+336x^2..... By equating coefficients: nb=28 and 0.5n(n-1)b^2=336 b^2=(28/n)^2 and b^2=336/0.5n(n-1) (28/n)^2=336/0.5(n-1). Solve to find n and hence b.