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Binomial theorem for positive integral Index

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Question

The sum of the rational terms in the expansion of $(\sqrt{2} + \sqrt[5]{3})^{10}$ , is

Moderate

Solution

Let $T_{r + 1}$ be $(r + 1)^{th}$ term in the expansion of
$(\sqrt{2} + \sqrt[5]{3})^{10}$ . Then
$T_{r + 1} =^{10} C_{r} (\sqrt{2})^{10 - r} (\sqrt[5]{3})^{r} =^{10} C_{r} \times 2^{5 - r / 2} \times 3^{r / 5}$
Here, $0 \leq r \leq 10$
For $T_{r + 1}$ to be rational
$\frac{r}{2}$ and $\frac{r}{5}$ must be integers.
$\Rightarrow$ $r$ must be a multiple of $10$ and $0 \leq r \leq 10$
$\Rightarrow r = 0, 10$
Thus, $T_{1}$ and $T_{11}$ are rational terms.
Now,
$T_{1} + T_{11} =^{10} C_{0} \times 2^{5} \times 3^{0} +^{10} C_{10} \times 2^{0} \times 3^{2} = 2^{5} + 9 = 41$

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