Let the coefficients of third, fourth and fifth terms in the expansion of (x+ax^2)^n, x0, …

Question

Let the coefficients of third, fourth and fifth terms in the expansion of $\left(x+\dfrac{a}{x^2}ight)^n$, $x
e0$, be in the ratio $12:8:3$. Then the term independent of $x$ in the expansion, is equal to

Answer: 4

Accepted Answer

$$T_{r+1}=\binom nr a^r x^{n-3r}$$

$$\frac{\text{coeff }T_3}{\text{coeff }T_4}=\frac{\binom n2a^2}{\binom n3a^3}=\frac{3}{a(n-2)}=\frac32\quad\Rightarrow\quad a(n-2)=2\quad\text{...(i)}$$

$$\frac{\text{coeff }T_4}{\text{coeff }T_5}=\frac{\binom n3a^3}{\binom n4a^4}=\frac{4}{a(n-3)}=\frac83\quad\Rightarrow\quad a(n-3)=\frac32\quad\text{...(ii)}$$

Solving (i),(ii): $n=6,\ a=\dfrac12$.

Independent term: $n-3r=0\Rightarrow r=2$: $T_3=\binom62\left(\frac12\right)^2=15\times\frac14=3.75\approx4$

Solutions · 1