The number of integral terms in the expansion of (3^(1/2) + 5^(1/4))⁶⁸⁰ is equal to

Q: Wait, why do we ignore the binomial coefficient ${}^{680}C_r$? Could it be a fraction?

Great observation! In the binomial expansion of $(a+b)^n$, the coefficient ${}^{n}C_r$ is always an integer for any integer $n$ and $0 \le r \le n$. Since we are dealing with integers, ${}^{680}C_r$ will always be an integer. Thus, the 'integral' nature of the term depends solely on the powers of the bases $3$ and $5$.

Q: Why does $r$ start at $0$ and go up to $680$? I thought there were $680$ terms.

Ah, the classic off-by-one error! For $(a+b)^n$, there are $n+1$ terms. Since $n=680$, there are $681$ terms in total. The index $r$ represents the power of the second term, so it ranges from $0$ to $n$. So, $r$ goes from $0$ to $680$.

Q: Why can't we just use energy conservation or some other physics trick?

Haha, I love the enthusiasm! But this is pure Algebra. In JEE, we must use the tools of the trade—Binomial Theorem. There is no 'energy' here, only the conservation of 'integrality'!

Question

The number of integral terms in the expansion of $(3^{\frac{1}{2}} + 5^{\frac{1}{4}})^{680}$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are examining the binomial expansion of $(3^{\frac{1}{2}} + 5^{\frac{1}{4}})^{680}$. Our objective is to determine the number of **integral terms** in this expansion.

An integral term is one where the resulting value is a pure integer, free from any irrational roots.

### The Master Key: The General Term

To solve this, we utilize the general term formula for a binomial expansion $(a+b)^n$, which is given by:

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

In this specific problem, we have $n=680$, $a=3^{\frac{1}{2}}$, and $b=5^{\frac{1}{4}}$. Substituting these values into the formula, we obtain:

$$T_{r+1} = {}^{680}C_r \cdot (3^{\frac{1}{2}})^{680-r} \cdot (5^{\frac{1}{4}})^r$$

Applying the laws of indices, specifically $(x^m)^n = x^{mn}$, the expression simplifies to:

$$T_{r+1} = {}^{680}C_r \cdot 3^{\frac{680-r}{2}} \cdot 5^{\frac{r}{4}}$$

### The Conditions of Existence

For $T_{r+1}$ to be an integer, the exponents of $3$ and $5$ must be non-negative integers. This leads us to two strict conditions:

1. $\frac{680-r}{2} \in \mathbb{Z}$
2. $\frac{r}{4} \in \mathbb{Z}$

For the first condition, $680-r$ must be divisible by $2$. Since $680$ is even, $r$ must be an even number.

For the second condition, $r$ must be a multiple of $4$. Because any number divisible by $4$ is inherently even, the second condition is the **dominant constraint**. We only need to ensure that $r$ is a multiple of $4$ within the range $0 \le r \le 680$.

### The Final Calculation

The values of $r$ that satisfy the condition are $r \in \{0, 4, 8, \dots, 680\}$. This sequence forms an **Arithmetic Progression**.

In this progression, the first term $a = 0$, the last term $l = 680$, and the common difference $d = 4$. The number of terms $N$ is calculated using the formula:

$$N = \frac{l-a}{d} + 1$$

Substituting our values into the equation:

$$N = \frac{680-0}{4} + 1 = 170 + 1 = 171$$

The total number of integral terms in the expansion is **171**.

Animated Solution for Mathematics - Binomial Theorem: The number of integral terms in the expansion of $(3^{\frac{1}{2}} + 5^{\frac{1}{4}})^{680}$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Expansion

The General Term Formula

Substituting the Values

Simplifying Exponents

Condition for Integral Terms

Analyzing the First Exponent

Analyzing the Second Exponent

Combining the Constraints

Forming the Sequence of $r$

Counting the Terms

Final Answer

The Sigma Insight: General Term and Middle Term

Analyzing the Setup

The Master Key

The Conditions of Existence

The Final Calculation

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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; } The number of integral terms in the expansion of (321​+541​)680 is equal to