The number of integral terms in the expansion of $\left( \frac{1}{5^2} + \frac{1}{7^8} \right)^{1016}$ is:

Step 1: Write the general term of the binomial expansion

We start by writing the general term of the binomial expansion $(a+b)^n$. In this problem, $a = \frac{1}{5^2}$, $b = \frac{1}{7^8}$, and $n = 1016$. The general term $T_{r+1}$ will help us identify which terms are integral.

Step 2: Substitute values into the general term

Now, we substitute the given values of $a$, $b$, and $n$ into the general term formula. This gives us the specific form of each term in the expansion.

Step 3: Simplify the general term

We simplify the expression by using the property $(x^m)^n = x^{mn}$ and $1/x^k = x^{-k}$. For a term to be integral, the powers of 5 and 7 must be non-negative integers. Since the powers are $2r-2032$ and $-8r$, for the term to be integral, both $2r-2032 \ge 0$ and $-8r \ge 0$ must hold.

Step 4: Determine the range of 'r' for integral terms

From the conditions derived in the previous step, we solve for $r$. The first inequality $2r - 2032 \ge 0$ implies $r \ge 1016$. The second inequality $-8r \ge 0$ implies $r \le 0$.

Step 5: Find common values of 'r'

We need to find values of $r$ that satisfy both $r \ge 1016$ and $r \le 0$. Since $r$ must also be an integer between $0$ and $n$ (inclusive), i.e., $0 \le r \le 1016$, there is no integer value of $r$ that can satisfy both conditions simultaneously.

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