Let us have a look at the t^{th} term.
Let us look at difference for a couple of terms
t_2 = t_1 = 5- 1 = 4
t_3-t_2 = 10 -5 = 5
t_4-t_3 = 16 -10 = 6
From the above we see that t_k = t_{k-1} + k + 2
From this let us find t_n in terms of n
We have t_n = 1 + \sum_{k=2}^{n} (k + 2)
= 1 + \frac{n(n+1)}{2} - 1 + 2(n-1)
= \frac{n^2}{2} + \frac{5n }{2} - 2
So sum S_n = \sum_{k=1}^n (\frac{k^2}{2} + \frac{5k }{2} - 2)
= \sum_{k=1}^n (\frac{k^2}{2}) + \sum_{k=1}^n\frac{5k }{2} - \sum_{k=1}^n2
=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6} + \frac{5}{2} \frac{n(n+1)}{2} - 2n
= \frac{2n^3 + 3n^2 +n }{12} + \frac{5n^2 +5n}{4} - 2n
= \frac{2n^3 + 3n^2 + n + 15n^2 + 15n -24n}{12}
= \frac{2n^3 + 18n^2 -8n}{12}
= \frac{n^3 + 9n^2 -4n}{6}
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