$1+ 4+10+ 20+35+ 56+ \ldots$ .

Find the sum of the above series up to $500^{th}$ term.

The answer is 2635531375.

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Call the terms in the series $a_1,a_2,\ldots$ . Note the difference $a_n-a_{n-1}$ is the $n$ -th triangular number, and $a_1$ is the first triangular number. Therefore, when summing the first $500$ terms, we are summing $501-n$ copies of the $n$ -th triangular number as $n$ ranges from $1$ to $500$ . Then the answer is $\sum_{k=1}^{500}\sum_{n=1}^{k}\frac{n(n+1)}{2}=\sum_{k=1}^{500}\frac{k(k+1)(k+2)}{6}=\frac{500\cdot 501\cdot 502\cdot 503}{24},$

i.e. the answer is $\boxed{2635531375}.$