Expansion (1+ax) ^n with the first term being 1 the second being 24x and the third term being 252x^2

please show workings with explanation

f(x) = (1 + ax)^n = 1 + 24x + 252x^2 + ...
f(0) = 1

f'(x) = an*(1+ax)^(n-1) = 24 + 504x + ...
f'(0) = an = 24

f''(x) = (a^2)n(n-1)*(1+ax)^(n-2) = 504 + ...
f''(0) = (a^2)n(n-1) = 504

So, our two equations are:

an = 24
(a^2)n(n-1) = 504

(a^2)(24/a)(24/a - 1) = 504

24a(24/a - 1) = 504

576 - 24a = 504

a = 3
n = 8

Thus, the original expression is:

(1 + 3x)^8

(1 + ax)^n

using binomial theorem

= 1 + nax +(1/2) n(n-1) a^2x^2 .........

=> 1 + nax + (1/2)(n^2 - n)a^2x^2 = 1 + 24x + 252x^2

comparing coefficients x and x^2 in LHS and RHS

na = 24 => a = 24/n

(1/2)(n^2 - n)a^2 = 252

substitute a = 24/n

=> (1/2)(n^2 - n)(24/n)^2 = 252

=> (n^2 - n)(576 /n^2) = 504

=> 576n^2 - 576n = 504n^2

=> 72n^2 = 576n

=> 72n = 576

n = 576/72

n = 8
a = 24/n = 3

a = 3 and n = 8

=> (1 + ax)^n = (1 + 3x)^8

= 1 + 8(3x) + 1/2(8*7)(9x^2) + 1/6(8*7*6)(27x^3)..............

= 1 + 24x + 252x^2 + 1512x^3 ..........

You need to find a and n to do expansion.
na = 24 => a = 24/n ......(1)
nC2 a^2 = 252 => n(n-1)/2 a^2 = 252 ......(2)
Plug (1) in (2), and simplify,
8(n-1) = 7n
n = 8
a = 3
(1+3x)^8 = 1 + 24x + 252x^2 + ... + (3x)^8

Learn Taylor's series.

Here is an excellent explanation: http://en.wikipedia.org/wiki/Taylor_seri…

The calculations are trivial.