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

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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**