Impossible Pattern

Let $\{a_n\} = 1, 2, 6, 15, 31, 56, ...$ such that $a_1=1$ , $a_2=2$ , $a_3=6$ and so on. We need to find $a_7$ . We note that:

$\begin{array} {ll} a_2-a_\red 1 = 2-1=1 & = \red 1^2 \\ a_3-a_\red 2 = 6-2=4 & = \red 2^2 \\ a_4-a_\red 3 = 15-6=9 & = \red 3^2 \end{array}$

$\begin{aligned} \implies a_{n+1}-a_{\color{#D61F06}n} & = {\color{#D61F06}n}^2 \\ a_7-a_{\color{#D61F06}6} & = {\color{#D61F06}6}^2 \\ \implies a_7 & = 36+56 = \boxed{92} \end{aligned}$

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If we want to find $a_n$ in terms of $n$ :

$\begin{aligned} a_{k+1} - a_k & = k^2 & \small \color{#3D99F6} \text{where }k \in \mathbb N \\ \sum_{k=1}^{n-1} a_{k+1} - \sum_{k=1}^{n-1} a_k & = \sum_{k=1}^{n-1} k^2 \\ a_n - a_1 & = \frac {n(n-1)(2n-1)}6 \\ \implies a_n & = \frac {n(n-1)(2n-1)}6 + 1 \\ a_7 & = \frac {7\cdot 6 \cdot 13}6 + 1 = \boxed{92} \end{aligned}$

92 FIRST TIME, we added 1 squared NEXT TIME, we added 2 squared and so on....

See this OEIS page.

formula: $(n+2)*(2*n^2-n+3)/6$

or you can accumulate squares: here is the $Mathematica$ code

1+Accumulate[Range@10^2]

which returns $2, 6, 15, 31, 56, 92, 141, 205, 286, 386...$