5, 10, 17, 26, 37, ...

I see the pattern, but I can't figure out the equation...any advice?

Printable View

- Mar 12th 2009, 10:53 PMjzelltNumber sequence
5, 10, 17, 26, 37, ...

I see the pattern, but I can't figure out the equation...any advice? - Mar 12th 2009, 11:29 PMu2_wa
Hello

**jzellt:**You might have noticed that difference between the numbers in sequence corresponds to the odd numbers sequence starting from 5.

$\displaystyle 5,7,9,11,.....$

Now we can write it as $\displaystyle 5+0,5+5,5+5+7,5+5+7+9,...$

We can use this formula for the addition of the odd numbers:

$\displaystyle \frac{n}{2}(2a+(n-1)*d)$

Lets plug in the values:

$\displaystyle \frac{n}{2}(2(5)+(n-1)*2)$

$\displaystyle \frac{n-1}{2}(10+(n-2)*2)$ (because at n=1, there should be addition of zero i.e**5+0**,5+5,....)

Hence we have the equation: $\displaystyle 5+\frac{n-1}{2}(8+2(n-1))$

Hope this helps - Mar 13th 2009, 04:09 AMSoroban
Hello, jzellt!

I "eyeballed" the solution . . .

Quote:

$\displaystyle 5, 10, 17, 26, 37, \hdots$

Take the differences of consecutive terms.

Then take the differences of the differences ... and so on.

. . $\displaystyle \begin{array}{cccccccccc}\text{Sequence} & 5 && 10 && 17 && 26 && 37 \\

\text{1st diff.} & & 5 && 7 && 9 && 11 \\

\text{2nd diff.} & & & 2 & & 2 & & 2 \end{array}$

We see that thedifferences are constant.*second*

This indicates that the generating function is of thedegree, a quadratic.*second*

. . That is, $\displaystyle f(n)$ contains $\displaystyle n^2.$

With that in mind, I looked the sequence again, and I saw:

. . $\displaystyle \begin{array}{c|ccc}

n & & a_n \\ \hline

1 & 5&=&2^2+1 \\

2 & 10 &=&3^2+1 \\

3 & 17 &=&4^2+1 \\

4 & 26 &=&5^2+1 \\

5 & 37 &=&6^2+1 \end{array}$

. . .$\displaystyle \begin{array}{c|c}\vdots & \vdots \\

n & \;\;(n+1)^2+1

\end{array}$

Therefore: .$\displaystyle f(n) \:=\:(n+1)^2+1 \:=\:n^2+2n+2$

. . which verifies u2_wa's result.