Let the coefficients of third- fourth and fifth terms in the expansion of x-ax2n- x- 0- be in the ratio 12-8-3- Then the term independent of x in the expansion- is equal to

Given : (x+ax^2)^n General term of the expansion is T_r+1=^nC_r x^n r(ax^2)^r ⇒ nbsp;T_r+1=^nC_ra^rx^n 3r Now, coefficient of T_3coefficient of T_4= 128 ⇒^nC ...

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A+B+C=N (N Fixed)

Let the coeff...

Question

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

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Solution

Given : ${(x + \frac{a}{x^{2}})}^{n}$
General term of the expansion is
$T_{r + 1} =^{n} C_{r} x^{n - r} {(\frac{a}{x^{2}})}^{r} \Rightarrow T_{r + 1} =^{n} C_{r} a^{r} x^{n - 3 r}$

Now,
$\frac{coefficient of T_{3}}{coefficient of T_{4}} = \frac{12}{8} \Rightarrow \frac{^{n} C_{2} \cdot a^{2}}{^{n} C_{3} \cdot a^{3}} = \frac{3}{2} \Rightarrow \frac{3}{a (n - 2)} = \frac{3}{2} \Rightarrow a (n - 2) = 2 \dots (i)$

Also,
$\frac{coefficient of T_{4}}{coefficient of T_{5}} = \frac{8}{3} \Rightarrow \frac{^{n} C_{3} \cdot a^{3}}{^{n} C_{4} \cdot a^{4}} = \frac{8}{3} \Rightarrow \frac{4}{a (n - 3)} = \frac{8}{3} \Rightarrow a (n - 3) = \frac{3}{2} \dots (i i)$
Using equations $(i)$ and $(i i)$ , we get
$n = 6, a = \frac{1}{2}$

Now, the term independent of $x$ is
$n - 3 r = 0 \Rightarrow r = 2 ∴ T_{3} =^{6} C_{2} {(\frac{1}{2})}^{2} = \frac{15}{4} = 3.75 = 4$

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Let the coefficients of third, fourth and fifth terms in the expansion of (x+ax2)n,x≠0, be in the ratio 12:8:3. Then the term independent of x in the expansion, is equal to

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