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

Let the general term in the binomial expansion of $$\left(5^{1/2}+7^{1/8}\right)^{1016}$$ be $$T_{k+1}$$, where $$k$$ ranges from $$0$$ to $$1016$$.

Using the Binomial Theorem,

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

Simplify the powers of $$5$$ and $$7$$:

$$T_{k+1} = \binom{1016}{k}\,5^{\frac{1016-k}{2}}\;7^{\frac{k}{8}}$$ $$-(1)$$

For $$T_{k+1}$$ to be an integer, the exponents of both $$5$$ and $$7$$ must be non-negative integers.

Condition 1 (for the power of 7)
The exponent $$\dfrac{k}{8}$$ must be an integer ⟹ $$k$$ must be a multiple of $$8$$.

Condition 2 (for the power of 5)
The exponent $$\dfrac{1016-k}{2}$$ must be an integer. Since $$1016$$ is even, $$\dfrac{1016-k}{2}$$ is an integer whenever $$k$$ is even.

If $$k$$ is a multiple of $$8$$, then $$k$$ is automatically even, so both conditions are satisfied simultaneously.

Hence, the permissible values of $$k$$ are all multiples of $$8$$ from $$0$$ up to $$1016$$ (inclusive):

$$k = 0,\,8,\,16,\,24,\,\dots,\,1016$$

The number of terms in this arithmetic sequence is

$$\text{Count} = \frac{1016}{8} + 1 = 127 + 1 = 128$$

Therefore, the expansion contains $$128$$ integral terms.

Option D is correct.