Question

The number of rational terms in the expansion of ${(\sqrt[3]{325}+\frac{1}{\sqrt[3]{325}})}^{20}$ is

• A. $2$
• B. $7$
• C. $6$
• D. $19$

Answer 1

Answer: (B)

$7$

$\text{Number of rational terms in}$

${(\sqrt[3]{325}+\frac{1}{\sqrt[3]{325}})}^{20}$

${{}^{20}(3\sqrt{25}{)}_{20}+{}^{20}{C}_1(3\sqrt{25})}^{19}\frac{1}{\sqrt[3]{325}}+\dots +{}^{20}{c}_{20}{\left(\frac{1}{\sqrt[3]{325}}\right)}^{20}$

Rational terms will be those where

$(3\sqrt{25}{)}^r{\left(\frac{1}{\sqrt[3]{325}}\right)}^{n-r}$ will be rational

and $(25{)}^{r/3}+\frac{r-n}{3}$

Therefore, $2r-n=3x;x$ is an integer

$n=20$

$x=\frac{2r-20}{3}\Rightarrow \frac{2(r-10)}{3}$

$\text{Putting values of}r,x\text{every}3\text{rd term is rational}$

$\text{there are}7\text{terms only;}$

$r=1,4,7,10,13,16,19$

$\text{So}7\text{terms are rational.}$