Answer

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**Hint:**We can see that the binomial expansion ${{\left( 1+ax \right)}^{n}}$ is similar to ${{\left( 1+x \right)}^{n}}$. So, we expand ${{\left( 1+ax \right)}^{n}}$ and compare the coefficients of x and ${{x}^{2}}$ on both sides to get the equations. We solve the equation to get the value of one of the variables. Using the obtained value of the variable, we find the value of the other variable by using one of the equations.

**Complete step by step answer:**

Given that we have a binomial expansion ${{\left( 1+ax \right)}^{n}}$, given as $1+8x+24{{x}^{2}}+......$. We need to find the values of ‘a’ and ‘n’.

The binomial expansion ${{\left( 1+ax \right)}^{n}}$ is similar to the expansion ${{\left( 1+x \right)}^{n}}$.

We know that the binomial expansion of ${{\left( 1+x \right)}^{n}}$ is given as:

${{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{3}}{{x}^{3}}+........$ ---(1).

We use equation (1), to expand the binomial expansion ${{\left( 1+ax \right)}^{n}}$.

So, the binomial expansion of ${{\left( 1+ax \right)}^{n}}$ is:

${{\left( 1+ax \right)}^{n}}=1+{}^{n}{{C}_{1}}\left( ax \right)+{}^{n}{{C}_{2}}{{\left( ax \right)}^{2}}+{}^{n}{{C}_{3}}{{\left( ax \right)}^{3}}+......$ ---(2).

According to the problem, the binomial expansion of ${{\left( 1+ax \right)}^{n}}$ is given as $1+8x+24{{x}^{2}}+......$ ---(3).

Using equations (2) and (3), we get

$1+8x+24{{x}^{2}}+......$ = $1+{}^{n}{{C}_{1}}\left( ax \right)+{}^{n}{{C}_{2}}{{\left( ax \right)}^{2}}+{}^{n}{{C}_{3}}{{\left( ax \right)}^{3}}+......$ .

$1+8x+24{{x}^{2}}+......$ = $1+{}^{n}{{C}_{1}}.a.x+{}^{n}{{C}_{2}}.{{a}^{2}}.{{x}^{2}}+{}^{n}{{C}_{3}}.{{a}^{3}}.{{x}^{3}}+......$ ---(4).

Comparing coefficients of ‘x’ terms on both sides, we get ${}^{n}{{C}_{1}}.a=8$.

We know that the value of ${}^{n}{{C}_{r}}$ is given as ${}^{n}{{C}_{r}}=\dfrac{n\times \left( n-1 \right)\times \left( n-2 \right)\times ..........\times \left( n-r+1 \right)}{r\times \left( r-1 \right)\times \left( r-2 \right)\times ..........\times 1}$.

So, we have got $\dfrac{n}{1}.a=8$.

We have got $n.a=8$ ---(5).

Equating coefficients of ${{x}^{2}}$ on both sides from equation (4), we get ${}^{n}{{C}_{2}}.{{a}^{2}}=24$.

We have got $\dfrac{n.\left( n-1 \right)}{1\times 2}.{{a}^{2}}=24$.

We have got $\dfrac{\left( {{n}^{2}}-n \right).{{a}^{2}}}{2}=24$.

We have got ${{n}^{2}}.{{a}^{2}}-n{{a}^{2}}=24\times 2$.

We have got ${{\left( n.a \right)}^{2}}-\left( n.a \right).a=48$ ---(6).

We substitute the value $n.a=8$ obtained from equation (5) in equation (6).

We have got ${{8}^{2}}-8a=48$.

We have got $64-48=8a$.

We have got $16=8a$.

We have got $a=\dfrac{16}{8}$.

We have got a = 2.

Now we substitute the obtained value of ‘a’ in the equation (5) to find the value of ‘n’.

We have got $n.2=8$.

We have got $n=\dfrac{8}{2}$.

We have got n = 4.

∴ The values of ‘a’ and ‘n’ is ‘2’ and ‘4’.

**So, the correct answer is “Option A”.**

**Note:**We need to make sure that the order of the answers given in options are for ‘a’ and ‘n’. Whenever we get the problems involving binomial expansion, we expand it first to solve. Similarly, we can expect to solve the problems that contain more than two variables and fractional powers.

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