Binomial expansion question (a-level maths)?

You exand the 16th power -- not the other. Then you start multiplying by the (23 + 13x) but only consider terms of the binomial expansion that produce x³ terms.

So (1 + x/2)^16 = C(n,0)(x/2)^0 + C(n,1)(x/2)^1 + C(n,2)(x/2)^2 + . . . etc. with n=16

Thus (23 +13x)(1 + x/2)^16 = (23 + 13x)(C(16,0)(x/2)^0 + C(16,1)(x/2)^1 + C(16,2)(x/2)^2 + C(16,3)(x/2)^3 . . . etc. ) =

(23 + 13x)(1 + x/2 + 16(x/2) + 120(x/2)^2 + 2135 (x/2)^3... etc. ) =

Now look at the products that produce cubes: constant times cube, linear times square:

23(2135)(x/2)^3 + 13x(120)(x/2)^2 =

6138.125 x^3 + 390 x^3 = 6528.125 x³

(Please check arithmetic in case I made an error,but that;s the idea.)

The first few terms of the binomial expansion of (2k + x)ⁿ are :- (2k)ⁿ + n(2k)ⁿ⁻¹x + n(n - 1)/2! *(2k)ⁿ⁻²x² + n(n - 1)(n - 2)/3! * (2k)ⁿ⁻³x³ +... a) As the coefficients of x² and x³ are the same then :- n(n - 1)/2! *(2k)ⁿ⁻² = n(n - 1)(n - 2)/3! * (2k)ⁿ⁻³ 2k =(n - 2)/3 : 6k = n - 2 : n = 6k + 2 b) As n = 6k + 2 then when k = ⅔, n = 6.....and using the earlier expansion gives :- (2*⅔)^6 + 6*(2*⅔)^5.x + 6*5/2! *(2*⅔)^4.x² + 6*5*4/3! * (2*⅔)³.x³ = (4096/729) + (2048/81)x + (1280/27)x² + (1280/27)x³........which is what you would expect as the coefficients of x² and x³ are equal.