If the first three terms of the expansion of [tex]$(1+ax)^n$[/tex] are [tex]$1+12x+64x^2$[/tex], find the values of [tex][tex]$n$[/tex][/tex] and [tex]$a$[/tex].



Answer :

To determine the values of [tex]\( n \)[/tex] and [tex]\( a \)[/tex] given the first three terms of the expansion of [tex]\( (1 + ax)^n \)[/tex], we start with the binomial expansion formula:

[tex]\[ (1 + ax)^n = \sum_{k=0}^{n} \binom{n}{k} (ax)^k \][/tex]

The first three terms of this expansion are:

[tex]\[ \binom{n}{0} (ax)^0 = 1, \quad \binom{n}{1} (ax)^1 = nax, \quad \binom{n}{2} (ax)^2 = \frac{n(n-1)}{2} a^2 x^2 \][/tex]

Given that the first three terms of the expansion are [tex]\( 1 + 12x + 64x^2 \)[/tex], we can match these coefficients with our expansion:

1. For the constant term:
[tex]\[ \binom{n}{0} (ax)^0 = 1 \quad \Rightarrow \quad 1 = 1 \][/tex]

2. For the coefficient of [tex]\( x \)[/tex]:
[tex]\[ \binom{n}{1} (ax)^1 = nax \quad \Rightarrow \quad na = 12 \][/tex]

3. For the coefficient of [tex]\( x^2 \)[/tex]:
[tex]\[ \binom{n}{2} (ax)^2 = \frac{n(n-1)}{2} a^2 x^2 \quad \Rightarrow \quad \frac{n(n-1)}{2} a^2 = 64 \][/tex]

Now, solve these equations step-by-step.

First, from [tex]\( na = 12 \)[/tex], we solve for [tex]\( a \)[/tex]:

[tex]\[ a = \frac{12}{n} \][/tex]

Next, substitute [tex]\( a \)[/tex] into the equation for the [tex]\( x^2 \)[/tex] term:

[tex]\[ \frac{n(n-1)}{2} \left( \frac{12}{n} \right)^2 = 64 \][/tex]

Simplify the equation:

[tex]\[ \frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 64 \quad \Rightarrow \quad \frac{144(n-1)}{2n} = 64 \quad \Rightarrow \quad \frac{72(n-1)}{n} = 64 \][/tex]

Multiply both sides by [tex]\( n \)[/tex]:

[tex]\[ 72(n-1) = 64n \quad \Rightarrow \quad 72n - 72 = 64n \quad \Rightarrow \quad 8n = 72 \quad \Rightarrow \quad n = 9 \][/tex]

Now, substitute [tex]\( n = 9 \)[/tex] back into the equation [tex]\( na = 12 \)[/tex]:

[tex]\[ 9a = 12 \quad \Rightarrow \quad a = \frac{12}{9} = \frac{4}{3} \][/tex]

Therefore, the values of [tex]\( n \)[/tex] and [tex]\( a \)[/tex] are:

[tex]\[ n = 9 \quad \text{and} \quad a = \frac{4}{3} \][/tex]