The number of integral terms in the expansion of ( 5^ 12 + 7^ 18)^1016 is:

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

Find: The count of terms that are integers.

Using the binomial theorem, the general term is

T_{k+1} = \binom{1016}{k} \left(5^{\frac{1}{2}}\right)^{1016-k} \left(7^{\frac{1}{8}}\right)^k

So,

T_{k+1} = \binom{1016}{k} 5^{\frac{1016-k}{2}} 7^{\frac{k}{8}}

For the term to be integral, both exponents must be integers.

\frac{1016-k}{2} \in \mathbb{Z}

which means $1016-k$ must be even, and

\frac{k}{8} \in \mathbb{Z}

which means $k$ must be a multiple of $8$ .

Since $1016$ is even, every multiple of $8$ automatically makes $1016-k$ even. Hence the allowed values are

k = 0, 8, 16, \dots, 1016

The valid values of k form an arithmetic progression with first term 0, common difference 8, and last term 1016. So the number of such values is 1016-08+1 = 127+1 = 128 Therefore, the number of integral terms is 128. Hence, the correct option is B.

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