<s>[INST] How to solve this sum problem? 
For the first radical section.
$$\sqrt{1\times 2\times 3\times 4 + 1} - 1 = 1 + 3 + 1 - 1= 5 - 1 = 4$$
The second radical section.
$$(\sqrt{2\times 3\times 4\times 5 + 1}) = 4 = 4 + 6 + 1 - 4 = 7$$
The third radical (not shown)
$$\sqrt{3\times 4\times 5\times 6 + 1} - 9 = 9 + 9 + 1 - 9 = 10$$
If $x$ is considered. Then:
$$\sqrt{x(x+1)(x+2)(x+3) + 1} - x^2 = 3x + 1$$
I need to find:
$$\sum_{n=1}^{97} 3n + 1$$
$$=3\sum_{n=1}^{97} n + 97$$
$$= 3\cdot 97 \cdot 49 + 97 = 14 356$$
 [/INST] Split your sum up:
$$3\sum\limits_{n=1}^{97}n + \sum\limits_{n=1}^{97}1$$
$$3 {97 \choose 2} + 97$$
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