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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 $a_{1}=2$?

Pizza Pizza
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  • Since the second difference is constant, there will be a quadratic solution. In fact $a_3=16+3$ and $a_6=49+3$ and $a_7= 65+1$ so the pattern you almost spotted is incorrect. – Henry Jun 09 '17 at 10:01
  • 2 10 19 29 40 52 65 79 94 110 127 145 164 184 205 227 250 274 299 325 $$a_1,a_2,,,,a_{20}$$ – Khosrotash Jun 09 '17 at 10:26
  • sequence of $a_n-a_{n-1}$ is (20 terms) $$\$$ 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 – Khosrotash Jun 09 '17 at 10:31

3 Answers3

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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 $a_n=a_{n-1}+n+6$

Emilio Novati
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Let $a_m=b_m+Am^2+Bm+C$

$2=a_1=b_1+A+B+C\iff b_1=?$

$$n+6=a_n-a_{n-1}=b_n-b_{n-1}+2An+B-A$$

Set $B-A=6,2A=1$ so that $b_n=b_{n-1}=\cdots=b_1=?$

lab bhattacharjee
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1

For $n \geq 2$ we have:

$$a_n =a_1 + \sum\limits_2^n (n+6) = a_1 + \sum\limits_2^n n + \sum\limits_2^n 6 $$

Can you take it from here?

EDIT: This result can help you with the first sum.

Piotr Benedysiuk
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