We distinguish between two cases for an admissible $50$-tuple $a_1, a_2, \ldots, a_{50}$.

a) If no $a_i$ is equal to $101$ then $a_1, a_2, \ldots, a_{50}$ contains exactly one number from every pair $(i, 101-i)$, $1 \le i \le 50$. Hence there are $2^{50}$ choices for $a_1, a_2, \ldots, a_{50}$, and each respective product $a_1a_2 \cdots a_{50}$ appears exactly once in the expansion of the product
$$P = (1 + 100)(2 + 99) \cdots (50 + 51) = 101^{50}.$$

b) If one of the $a_i$ is $101$ then the remaining ones come from $49$ different pairs $(i, 101-i)$, $1 \le i \le 50$. Suppose that pair $(1, 100)$ is not present. There are $2^{49}$ such products $a_1a_2 \cdots a_{50}$, the summands in the expansion of the $P_1 = 101(2+99)\cdots(50+51) = 101^{50}$. Analogously if the non-represented pair is $(2, 99)$, $(3, 98)$, \dots, $(50, 51)$ the respective products appear once in the expansions of
$$\begin{align*} P_2 &= (1 + 100)101(3 + 98)\cdots(50 + 51) = 101^{50}, \ P_3 &= (1 + 100)(2 + 99)101\cdots(50 + 51) = 101^{50}, \\ \multicolumn{2}{l}{\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots} \\ P_{50} &= (1 + 100)(2 + 99)\cdots(49 + 52)101 = 101^{50}. \end{align*}$$
By a) and b) the sum in question is $101^{50} + 50 \cdot 101^{50} = 51 \cdot 101^{50}$.