discrete mathematics - Find closed form for $a_{1}=2, a_{n}=a_{n-1}+n+6$

I have determined that $a_{2} = 10, a_{3} = 19, a_{4} = 29, a_{5} = 40, a_{6} = 52,$ and $a_{7} = 65$. I can see that there is a pattern in that each value increases by 8, then 9, then 10, then 11, then 12, etc. but I am having difficulty making an equation for it.

I thought I had it when I realized that $a_{2} = 9+1, a_{3} = 16 + 4, a_{4} = 25 + 4, a_{5} = 36 + 4,$ and so on, but then I realized it was not very consistent. Also, the difference between $a_{n} - (n-1)^2$ starts to get smaller as the value of n increases, and then begins to increase again later.

Am I going about this completely wrong? Is there a way to find a closed form for $a_{n}=a_{n-1}+n+6$ when $a_{1}=2$?

Answer

Hint:

reorder the terms:

$$a_1=2$$
$$a_2=a_1+2+6=2+2+6$$
$$a_3=a_2+3+6=2+2+6+3+6=2+(2+3)+2\cdot 6$$
$$a_4=a_3+4+6=2+(2+3)+2\cdot 6+4+6=2+(2+3+4)+3\cdot 6$$

this suggests:

$$a_n=2+\frac{(n-1)(2+n)}{2}+6(n-1)$$

Now prove that it works giving $a_n=a_{n-1}+n+6$