7. A number table composed of several rows of numbers, starting from the second row, each number in the row is equal to the sum of the two numbers above it, the last row contains only one number, the first row is the first 100 positive integers arranged in ascending order, then the number in the last row is $\qquad$ (exponential notation can be used).





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The numbers in the $n$-th row form an arithmetic sequence with a common difference of $2^{n-1}$. Let the first number in the $n$-th row be $a_{n}$, then $a_{n+1}=2 a_{n}+2^{n-1} \Rightarrow \frac{a_{n+1}}{2^{n-1}}=\frac{a_{n}}{2^{n-2}}+1 \Rightarrow\left\{\frac{a_{n}}{2^{n-2}}\right\}$ is an arithmetic sequence with a common difference of 1. Also, $\frac{a_{1}}{2^{-1}}=2 \Rightarrow \frac{a_{n}}{2^{n-2}}=n+1 \Rightarrow a_{n}=(n+1) \cdot 2^{n-2}\left(n \in \mathbf{N}^{*}\right)$.
Therefore, the number in the last row is $a_{100}=101 \cdot 2^{98}$.