The sum of all rational terms in the expansion of $\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$ is equal to :
Solution
<p>$$\begin{aligned}
& T_{r+1}={ }^{15} \mathrm{C}_r\left(2^{1 / 5}\right)^{15-r}\left(5^{1 / 3}\right)^r \\
& ={ }^{15} C_r 5^{r / 3} 2^{\left(3-\frac{r}{5}\right)}
\end{aligned}$$</p>
<p>For rational terms,</p>
<p>$\frac{r}{3}$ and $\frac{r}{5}$ must be integer</p>
<p>3 and 5 divide $r \Rightarrow 15$ divides $r \Rightarrow r=0$ and $r=15$</p>
<p>${ }^{15} C_0 5^0 2^3+{ }^{15} C_{15} 5^5 2^{(0)}$</p>
<p>$$\begin{aligned}
& =8+3125 \\
& =3133
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
This question is part of PrepWiser's free JEE Main question bank. 193 more solved questions on Binomial Theorem are available — start with the harder ones if your accuracy is >70%.