Given that the binomial expansion of (1+kx)^n begins 1+8x+16x^2+... a) find k and n b) for what x is this expansion valid?

a) We compare the expansion given to the standard binomial expansion (remembering the powers of k).(1+kx)ⁿ=1+n(kx)+(n(n-1)/2)(kx)²+...As this is true for all x (for which the expansion holds), we can compare coefficients. So nk=8 and k²n(n-1)/2=16 (or k²n(n-1)=32).Then we can solve these simultaneous equations by substitution. Rearrange the first equation to obtain k=8/n. Then substitute this into the second equation to obtain (8/n)²n(n-1)=32. Rearrange to obtain 2(n-1)/n=1, and then obtain n=2. Substitute this into k=8/n to get k=4.b) Now we require |kx|<1 for the expansion to hold, and as we now know k=4, we must have |x|<1/4.