An algebra problem by Vilakshan Gupta

We note that:

$\begin{aligned} t_2-t_1 & = 3-2 = 1 \\ t_3-t_2 & = 6-3 = 3 \\ t_4-t_3 & = 11-6 = 5 \\ t_5-t_4 & = 18-11 = 7 \\ \implies t_{n+1} - t_n & = 2n - 1 \\ \implies \sum_{\color{#3D99F6}n=1}^{\color{#3D99F6}m} t_{\color{#3D99F6}n+1} - \sum_{n=1}^m t_n & = \sum_{n=1}^m (2n-1) \\ \sum_{\color{#D61F06} n=2}^{\color{#D61F06}m+1} t_{\color{#D61F06}n} - \sum_{n=1}^m t_n & = 2 \sum_{n=1}^m n - \sum_{n=1}^m 1 \\ t_{m+1} - t_1 & = n(n+1) - n \\ \implies t_{m+1} & = n^2 + t_1 = n^2 + 2 \\ \implies t_{50} & = 49^2 + 2 = \boxed{2403} \end{aligned}$

2,3,6,11,18

Can be rewritten as

2+0, 2+(1^2) , 2+(2^2) , 2+(3^2) , 2+(4^2)

2+(n-1)^2

Therefore,tn=2+(n-1)^2

So, t50 = 2+(49)(49) =2403