Question

In the expansion of $(1+ax{)}^n$, $n\in N$, then the coefficient of x and $x^2$ are $8$ and $24$ respectively. Then?

• A. $a=2,n=4$
• B. $a=4,n=2$
• C. $a=2,n=6$
• D. None of these

Answer 1

The correct option is A $a=2,n=4$

Here in the expansion of $(1+x.a{)}^n$,

$n\in N$, the $(r+1{)}^{th}$ term will be

$T_{r+1}=\left(\frac{n}{r}\right)a^{n-r}.b^r$

here first term $a=1$ &

second term $b=x.a$, so,

$T_{r+1}=\left(\frac{n}{r}\right)(1{)}^{n-r}.(x.a{)}^r=\left(\frac{n}{r}\right).x^r.a^r$

But given that, coefficient of $x$ is $8$ so,

For $x$ power $1$, we have to put $r=1$

So, $T_2=\left(\frac{n}{1}\right)a^1.x^1$

So, $\left(\frac{n}{1}\right)a=8\to \left(i\right)$

Now, coefficient of $x^2$ is $24$, so, $r=2$

$T_3=\left(\frac{n}{2}\right)a^2.x^2$

So, $24=\left(\frac{n}{2}\right)a^2\to \left(ii\right)$

From $\left(1\right)$ & $\left(2\right),a.n=8\Rightarrow a=8/n$

$\frac{n(n-1)}{2}.a^2=24$

$\frac{n(n-1)}{2}{\left(\frac{8}{n}\right)}^2=24$

$\therefore \frac{n^2-n}{n^2}\times 32=24$

$\therefore 1-\frac{1}{n}=\frac{24}{32}=\frac{3}{4}$, So, $n=4$ from $\left(i\right)$

$a=2$