Question

In the expansion of ${}^{20}{C}_r$.$4^{\frac{20-r}{3}}{.6}^{-\frac{r}{4}}$

• A. the total no. of rational terms = 4
• B. the total no.of irrational terms = 18
• C. the middle term is rational
• D. the total no.of irrational terms = 19

Answer 1

The correct option is D

the total no.of irrational terms = 19

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

= ${}^{20}{C}_r$. $2^{\frac{40-2r}{3}}{.2}^{-\frac{r}{4}}{.3}^{-\frac{r}{4}}$

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

For rational terms, $\frac{r}{4}$ and $\frac{160-11r}{12}$ must be integers and 0 ≤ r ≤ 20.

So, $\frac{r}{4}$ = integer ⇒ r = 0, 4, 8, 12, 16, 20

∴ Only two terms are rational.

So 21 - 2 i.e. 19 terms are irrational.