Question

Let the coefficients of third, fourth and fifth terms in the expansion of ${\left[x+\left(\frac{a}{x^2}\right)\right]}^n,x\ne 0$ be in the ratio $12:8:3$. Then the term independent of $x$ in the expansion is equal to ………

Answer 1

Step 1: Finding the term independent of $x$ using Binomial expansion

The given expression

${\left[x+\left(\frac{a}{x^2}\right)\right]}^n,x\ne 0$

So, ${(r+1)}^{th}$ of this expression

$$\begin{gathered} \Rightarrow T_{r+1}={}^n{C}_r{x}^{n-r}{\left(\frac{a}{x^2}\right)}^r \\ ={}^n{C}_r{x}^{n-3r}{a}^r...\left(i\right) \\ \Rightarrow T_{2+1}={}^n{C}_2{x}^{n-6}{a}^2\mathbf{}\mathbf{[}\mathbf{}\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{]} \\ \Rightarrow T_{3+1}={}^n{C}_3{x}^{n-9}{a}^3\mathbf{[}\mathbf{}\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}] \\ \Rightarrow T_{4+1}={}^n{C}_4{x}^{n-12}{a}^4\mathbf{[}\mathbf{}\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{]} \end{gathered}$$

Step 2: Dividing coefficients of $T_3\&T_4$

$\frac{[T_3]}{[T_4]}=\frac{\left[{}^n{C}_2{a}^2\right]}{\left[{}^n{C}_3{a}^3\right]}=\frac{12}{8}$

so,

$$\begin{gathered} \Rightarrow \frac{\left[{\displaystyle \frac{n(n-1)}{2}}{a}^2\right]}{\left[\frac{n(n-1)(n-2)}{3\times 2}{a}^3\right]}=\frac{3}{2} \\ \Rightarrow \frac{3}{a(n–2)}=\frac{3}{2} \\ \Rightarrow a(n–2)=2...\left(ii\right) \end{gathered}$$

Step 3: Dividing coefficients of $T_4\&T_5$

$\frac{[T_4]}{[T_5]}=\frac{\left[{}^n{C}_3{a}^3\right]}{\left[{}^n{C}_4{a}^4\right]}=\frac{8}{3}$

so,

$$\begin{gathered} \Rightarrow \frac{\left[\frac{n(n-1)(n-2)}{3\times 2}{a}^3\right]}{\left[\frac{n(n-1)(n-2)(n-3)}{4\times 3\times 2}{a}^4\right]}=\frac{8}{3} \\ \Rightarrow a(n–3)=\frac{3}{2}...\left(iii\right) \end{gathered}$$

By solving $\left(i\right)$ and $\left(ii\right)$ we have

$n=6\&\hspace{0.17em}a=\frac{1}{2}$

Step 4: Calculating independent term of ‘$x$,

Substituting,

$$\begin{gathered} n–3r=0\mathbf{}\mathbf{\left[}\mathbf{\text{from equation (i}}\mathbf{)}\mathbf{\right]} \\ \Rightarrow r=\frac{n}{3} \\ =\frac{6}{3} \\ =2 \\ \therefore T_3={}^6{C}_2{\left(\frac{1}{2}\right)}^2{x}^0\left[\because n=6\&a=\frac{1}{2}\right] \\ =\frac{15}{4} \\ =3.75\cong 4 \end{gathered}$$

Hence, the term independent of x in the expansion is equal to 3rd term and value is $3.75$