National Olympiad of Argentina

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

a) If no $a_i$ is equal to $101$ then $a_1,a_2,\dots ,a_{50}$ contains exactly one number from every pair $(i,101-i)$, $1\le i\le 50$. Hence there are $2^{50}$ choices for $a_1,a_2,\dots ,a_{50}$, and each respective product $a_1{a}_2\cdots a_{50}$ 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 $a_1{a}_2\cdots a_{50}$, 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 $$\begin{array}{rlr}{P}_2& =(1+100)101(3+98)\cdots (50+51)=101^{50},P_3& =(1+100)(2+99)101\cdots (50+51)=101^{50},\\ \text{multicolumn}2l\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots & & \\ P_{50}& =(1+100)(2+99)\cdots (49+52)101=101^{50}.& \end{array}$$ By a) and b) the sum in question is $101^{50}+50\cdot 101^{50}=51\cdot 101^{50}$.