In the expansion of (1+ax)ⁿ, the first three terms are: 1+24x+252x². Find a and n.Binomial Theorem chapter.

The binomial theorem gives the first three terms of the
expansion of (1 + ax)^n as 1, [n!/1!*(n - 1)!]*ax and [n!/2!*(n -
2)!]a^2x^2

From the information
given:

[n!/1!*(n - 1)!]*ax = 24x and [n!/2!*(n - 2)!]a^2x^2
= 252x^2

=> n*a*x = 24x and [n(n - 1)/2]*a^2*x^2 =
252x^2

=> na = 24 and [n(n - 1)/2]*a^2 =
252

a = 24/n

substitute in
[n(n - 1)/2]*a^2 = 252

=> [n(n - 1)*288] =
252n^2

=> (n^2 - n)*8 =
7n^2

=> 8n^2 - 7n^2 =
8n

=> n^2 - 8n =
0

=> n = 0 and n = 8

a
= inf.  and a = 3

As a = inf. is not possible take only the
solution a = 3 and n = 8

The required values
are a = 3 and n = 8