Understanding the Key Concept: The Binomial Theorem and Integral Terms

The problem asks us to find the number of integral terms in the expansion of an expression of the form $(a^{1/p} + b^{1/q})^n$. To solve this, we first need to recall the Binomial Theorem.

The Binomial Theorem states that the general term (or $(r+1)^{th}$ term) in the expansion of $(X+Y)^n$ is given by: $T_{r+1} = \binom{n}{r} X^{n-r} Y^r$ where $r$ can take integer values from $0$ to $n$.

For a term to be integral, two conditions must be met:

1. The binomial coefficient $\binom{n}{r}$ must be an integer (which is always true when $n$ and $r$ are integers with $0 \le r \le n$).
2. The powers of the base numbers (e.g., $a$ and $b$ in our case) must result in integers. This means if the base numbers are themselves integers, their exponents must be non-negative integers.

Step-by-Step Solution

1. Write Down the General Term

Given the expression $\left( {5^\frac{1}{2}} + 7^\frac{1}{8} \right)^{1016}$, we can identify:

• $X = 5^\frac{1}{2}$
• $Y = 7^\frac{1}{8}$
• $n = 1016$

Using the general term formula, $T_{r+1} = \binom{n}{r} X^{n-r} Y^r$, we substitute these values: $T_{r+1} = \binom{1016}{r} \left(5^\frac{1}{2}\right)^{1016-r} \left(7^\frac{1}{8}\right)^r$ This simplifies to: $T_{r+1} = \binom{1016}{r} 5^{\frac{1016-r}{2}} 7^{\frac{r}{8}}$ Explanation: This step is crucial because it gives us the general form of any term in the expansion. By analyzing this term, we can establish conditions for it to be an integer. The value of $r$ can range from $0$ to $1016$, inclusive, i.e., $0 \le r \le 1016$.

2. Establish Conditions for an Integral Term

For $T_{r+1}$ to be an integral term, the exponents of $5$ and $7$ must both be non-negative integers. Since $5$ and $7$ are integers, their powers must also be integers for the term to be integral.

From $5^{\frac{1016-r}{2}}$, the exponent $\frac{1016-r}{2}$ must be an integer. This implies that $(1016-r)$ must be divisible by $2$. Since $1016$ is an even number, $(1016-r)$ will be divisible by $2$ if and only if $r$ is an even number. So, Condition 1: $r$ must be an even integer.

From $7^{\frac{r}{8}}$, the exponent $\frac{r}{8}$ must be an integer. This implies that $r$ must be divisible by $8$. So, Condition 2: $r$ must be a multiple of $8$.

Explanation: We need to ensure that the fractional powers disappear, leaving only integer powers of the prime bases (5 and 7). This directly leads to the divisibility conditions for $r$.

3. Combine Conditions and Find Possible Values of $r$

Combining Condition 1 ($r$ is an even integer) and Condition 2 ($r$ is a multiple of $8$), we conclude that $r$ must be a multiple of $8$. (If $r$ is a multiple of 8, it is automatically an even number).

Also, we know that $r$ must satisfy $0 \le r \le 1016$.

So, the possible values of $r$ are multiples of $8$ within this range: $r \in \{0, 8, 16, 24, \ldots, 1016\}$

Explanation: The range of $r$ is defined by the binomial expansion. We are looking for values of $r$ that satisfy both conditions for integrality within this valid range.

4. Count the Number of Such Terms using Arithmetic Progression

The values of $r$ form an Arithmetic Progression (AP) with:

• First term, $a = 0$
• Common difference, $d = 8$
• Last term, $l = 1016$

Let $N$ be the number of terms in this AP. The formula for the $N^{th}$ term of an AP is $l = a + (N-1)d$. Substituting the values: $1016 = 0 + (N-1)8$ $1016 = 8(N-1)$

Divide by $8$: $\frac{1016}{8} = N-1$ $127 = N-1$

Solving for $N$: $N = 127 + 1$ $N = 128$

Therefore, there are $128$ integral terms in the expansion.

Explanation: Once we have a sequence of values for $r$ that are equally spaced, an arithmetic progression is the most efficient way to count them. Each valid value of $r$ corresponds to exactly one integral term in the expansion.

Tips and Common Mistakes:

• Don't forget $r=0$: The first term corresponds to $r=0$, which is often integral and should not be overlooked.
• Range of $r$: Always remember that $0 \le r \le n$. Terms outside this range are not part of the expansion.
• Least Common Multiple: If the exponents required $r$ to be a multiple of, say, 2 and 3, then $r$ would need to be a multiple of LCM(2,3) = 6. In this problem, being a multiple of 8 automatically satisfies being an even number, so LCM(2,8) = 8.
• Arithmetic Series vs. Number of Terms: Be careful when using AP formulas. We are counting the number of terms, $N$, not the sum of the terms.

Summary and Key Takeaway:

To find the number of integral terms in a binomial expansion of the form $(a^{1/p} + b^{1/q})^n$:

1. Write out the general term $T_{r+1} = \binom{n}{r} a^{\frac{n-r}{p}} b^{\frac{r}{q}}$.
2. Determine the conditions on $r$ such that $\frac{n-r}{p}$ and $\frac{r}{q}$ are both non-negative integers. This typically involves $r$ being a multiple of certain numbers.
3. Combine these conditions with the range $0 \le r \le n$.
4. Use the formula for the number of terms in an arithmetic progression to count the valid values of $r$. Each valid $r$ corresponds to one integral term.

The final answer is $\boxed{\text{128}}$.