Number of integral terms in the expansion of {7^(1/2) + 11^(1/6)}⁸²⁴ is equal to_____

Q: Wait, why does $r$ start at $0$? Shouldn't it start at $1$?

Great question! In the binomial expansion of $(a+b)^n$, the general term is denoted as $T_{r+1}$. When $r=0$, we get the first term $T_1$. If we started at $r=1$, we would be looking at the second term. Always remember that $r$ ranges from $0$ to $n$.

Q: Why do we only care about the exponents of $7$ and $11$? What about the binomial coefficient $\binom{824}{r}$?

That is a brilliant observation! The binomial coefficient $\binom{n}{r}$ is defined as $\frac{n!}{r!(n-r)!}$, which is always an integer for any valid $r$. Since the product of integers is an integer, we only need to ensure the powers of the prime bases $7$ and $11$ are integers to make the whole term an integer.

Q: What if the denominators were not multiples of each other? Like $3$ and $4$?

The logic remains identical! You would still find the LCM of the denominators. If the denominators were $3$ and $4$, $r$ would need to be a multiple of $LCM(3, 4) = 12$. The principle of finding the LCM of the denominators is the universal key to solving these types of problems.

Question

Number of integral terms in the expansion of $\{7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}\}^{824}$ is equal to_____

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, intimidating expression: $(7^{\frac{1}{2}} + 11^{\frac{1}{6}})^{824}$. It looks like a mountain of irrational numbers, but in the world of JEE Advanced, we don't fear the mountain; we climb it.

We are going to find the 'integral terms' hidden within this expansion. Think of this as a treasure hunt where we are looking for the rare, perfect integers hiding in a sea of radicals.

### The General Term: Our Compass

To navigate this, we use the General Term Formula. For any binomial expansion $(a + b)^n$, the $(r+1)$-th term is given by:

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

Here, our $a = 7^{\frac{1}{2}}$, our $b = 11^{\frac{1}{6}}$, and our $n = 824$. Plugging these into the formula, we get:

$$T_{r+1} = \binom{824}{r} (7^{\frac{1}{2}})^{824-r} (11^{\frac{1}{6}})^r$$

Using the exponent rule $(x^m)^n = x^{mn}$, we simplify this into:

$$T_{r+1} = \binom{824}{r} 7^{\frac{824-r}{2}} 11^{\frac{r}{6}}$$

### The Integral Constraint

Now, we want $T_{r+1}$ to be an integer. The binomial coefficient $\binom{824}{r}$ is always an integer, so the real challenge lies in the exponents of $7$ and $11$.

For the term to be an integer, the exponents $\frac{824-r}{2}$ and $\frac{r}{6}$ must be non-negative integers. This gives us two conditions:

1. $824-r$ must be divisible by $2$.
2. $r$ must be divisible by $6$.

Since $824$ is even, $824-r$ being divisible by $2$ simply means $r$ must be even. Because any multiple of $6$ is automatically even, the condition simplifies beautifully: **$r$ must be a multiple of $6$**.

### The Arithmetic Progression

We are looking for values of $r$ such that $r = 6k$, where $k$ is an integer. Given the constraint $0 \le r \le 824$, we determine the range of $k$.

The smallest value is $r=0$ (when $k=0$). To find the largest value, we divide $824$ by $6$:

$$824 = 6 	imes 137 + 2$$

Thus, the largest valid $r$ is $824 - 2 = 822$. Our sequence of $r$ values is $0, 6, 12, \dots, 822$.

This is an Arithmetic Progression where the first term $a = 0$, the common difference $d = 6$, and the last term $l = 822$. Using the formula $l = a + (n-1)d$:

$$822 = 0 + (n-1)6$$

Solving for $n$:

$$n-1 = \frac{822}{6} = 137 \Rightarrow n = 138$$

### The Victory

We have systematically broken down the constraints to find the number of integral terms. There are exactly **$138$** terms in this expansion that are integers.

It is not just about the calculation; it is about seeing the structure beneath the chaos. You have mastered the logic of binomial constraints.

Animated Solution for Mathematics - Binomial Theorem: Number of integral terms in the expansion of ${7^{(\frac{1}{2})} + 1 1^{(\frac{1}{6})}}^{824}$ is equal to_____

Enter Numerical Value:

Visualized Solution

The General Term Formula

Substituting the Values

Simplifying the Exponents

The Condition for Integral Terms

Analyzing the First Exponent

Analyzing the Second Exponent

Combining the Constraints

Identifying the Sequence

Setting up the Arithmetic Progression

Calculating the Number of Terms

Final Answer

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The General Term

The Integral Constraint

The Arithmetic Progression

The Victory

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of integral terms in the expansion of {7(21​)+11(61​)}824 is equal to_____