Question

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

Hint:

Use the ratios of consecutive coefficients \(\binom{n}{2}a^2 : \binom{n}{3}a^3 : \binom{n}{4}a^4\) to find n and a, then set the power of x to zero.

Answer 1

Correct Answer: 3.75

The coefficients of the 3rd, 4th, 5th terms are \(\binom{n}{2}a^2,\ \binom{n}{3}a^3,\ \binom{n}{4}a^4\).
Ratio of 4th to 3rd: \(\frac{\binom n3 a}{\binom n2} = \frac{(n-2)a}{3} = \frac{8}{12}\), so \((n-2)a = 2\).
Ratio of 5th to 4th: \(\frac{\binom n4 a}{\binom n3} = \frac{(n-3)a}{4} = \frac38\), so \((n-3)a = \frac32\).
Subtracting: \(a = \frac12\), and then \(n - 2 = 4\), so \(n = 6\).
General term: \(T_{r+1} = \binom6r x^{6-r}\left(\frac{a}{x^2}\right)^r = \binom6r a^r x^{6-3r}\). Independent of x when \(6 - 3r = 0\), so \(r = 2\).
Term \(= \binom62\cdot\left(\frac12\right)^2 = \frac{15}{4} = 3.75\).
Answer: 3.75