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So what answer would it be using that?

so in general to find the (n+1)th term from the nth you do:
\[\frac{n(n+1)(2n+1)}{6}+(n+1)^2\]

where n(n+1)(2n+1)/6 was the nth term

and (n+1)^2 was what you needed to add to get the next number (that I called (n+1)th

so mathstudent is right?

yes (99+1)^2

I don't think he ever wrong :p

he was right, thank you guys