Question

In the expansion of ${(4^{\frac{1}{3}}+6^{\frac{-1}{4}})}^{20}$, consider the following statements:

(i) the number of irrational terms = 18

(ii) the number of rational terms = 2

(iii) the middle term is irrational.

Then the correct statements are:

• A. (i) and (iii)
• B. (ii) and (iii)
• C. (i) and (ii)
• D. All three

Answer 1

The correct option is B

(ii) and (iii)

$T_{r+1}={}^{20}{C}_r{4}^{\frac{(20-r)}{3}}(6{)}^{\frac{-r}{4}}$

= ${}^{20}{C}_r{2}^{\frac{(160-11r)}{12}}{3}^{\frac{-r}{4}}$.

For rational terms, (160 - 11r) and $\frac{r}{4}$ must be integers and $0\le r\le 20$

$\therefore \frac{r}{4}=integerr=0,4,8,12,16,20ofthese,$ r = 8, 20 give $\frac{(160-11r)}{12}$ also an integer. Therefore,only two terms are rational and so, 21 - 2 = 19 terms are irrational.

Mid. term = $T_{10+1}$ is irrational.