Let the coefficients of third, fourth and fifth terms in the expansion of $\left(x+\frac{a}{x^{2}}\r

Question

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

Professor PC Thomas Classes · Answer

(4) : $$\begin{equation*}T_{r+1}={ }^{n} C_{r}(x)^{n-r}\left(\frac{a}{x^{2}}ight)^{r}={ }^{n} C_{r} a^{r} x^{n-3 r} 	ag{i} \end{equation*}$$ We have, $\frac{{ }^{n} C_{2} a^{2}}{{ }^{n} C_{3} a^{3}}=\frac{12}{8} \Rightarrow 2=a(n-2)$\ and $\frac{{ }^{n} C_{3} a^{3}}{{ }^{n} C_{4} a^{4}}=\frac{8}{3} \Rightarrow 3=2 a(n-3)$\ Dividing (i) by (ii), we get\ $\frac{2}{3}=\frac{(n-2)}{2(n-3)} \Rightarrow n=6$\ From (i), $2=a(6-2) \Rightarrow a=\frac{1}{2}$\ For the term independent of $x, n-3 r=0$\ $\Rightarrow 6-3 r=0 \Rightarrow r=2$\ So, $T_{3}={ }^{6} C_{2} a^{2}=\frac{15}{4} \approx 4$ (Approx.)\