Animated Solution for Mathematics - Binomial Theorem: The sum of the rational terms in the expansion of $(\sqrt{2}+3^{1/5}{)}^{10}$ is .........

A Quest for Rationality

Imagine you are standing before the expression $(\sqrt{2}+3^{1/5}{)}^{10}$. It looks innocent enough, but it is a gateway to a hidden structure.

When we expand this using the Binomial Theorem, we are essentially creating a collection of eleven distinct terms. Most of these terms will be tangled in the roots of 2 and 3, but a few—the 'rational' ones—will emerge as clean, whole numbers.

Our mission is to find these hidden gems and sum them up.

The General Term

We do not need to expand the entire expression to find these terms. Instead, we use the general term formula for $(a+b{)}^n$, which is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$.

This formula is our telescope; it allows us to zoom in on any term $T_{r+1}$ without looking at the rest. Here, our $a=2^{1/2}$, $b=3^{1/5}$, and $n=10$.

Substituting these into our formula, we get:

Simplifying the exponents, we arrive at the master expression:

Now, we apply the filter. For a term to be rational, it must be free of any fractional powers. This means the exponents of our prime bases—2 and 3—must be integers.

We have two conditions to satisfy:

1. The exponent of 2 must be an integer: $\frac{10-r}{2}\in \mathbb{Z}$. 2. The exponent of 3 must be an integer: $\frac{r}{5}\in \mathbb{Z}$.

Since $r$ represents the index of the term, it must be an integer such that $0\le r\le 10$.

Let us look at the second condition first: $\frac{r}{5}$ is an integer only if $r$ is a multiple of 5. Within our range of $0$ to $10$, the candidates for $r$ are $0,5,$ and $10$.

Now, we test our candidates against the first condition, $\frac{10-r}{2}\in \mathbb{Z}$:

For $r=0$: $\frac{10-0}{2}=5$. This is an integer! $r=0$ is a winner. For $r=5$: $\frac{10-5}{2}=2.5$. This is not an integer. We must reject $r=5$. For $r=10$: $\frac{10-10}{2}=0$. This is an integer! $r=10$ is a winner.

With our valid $r$ values identified, we calculate the terms:

For $r=0$:

For $r=10$:

The rational terms are $32$ and $9$. Adding them together, we get $32+9=41$.

We have navigated the binomial landscape and found our treasure. The sum is 41.