Number sequence

**jzellt** · March 12th 2009, 10:53 PM

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

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

**u2_wa** · March 12th 2009, 11:29 PM

Originally Posted by jzellt

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

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

Hello jzellt:

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

$5,7,9,11,.....$

Now we can write it as $5+0,5+5,5+5+7,5+5+7+9,...$
We can use this formula for the addition of the odd numbers:
$\frac{n}{2}(2a+(n-1)*d)$

Lets plug in the values:
$\frac{n}{2}(2(5)+(n-1)*2)$
$\frac{n-1}{2}(10+(n-2)*2)$ (because at n=1, there should be addition of zero i.e 5+0,5+5,....)
Hence we have the equation: $5+\frac{n-1}{2}(8+2(n-1))$

Hope this helps

**Soroban** · March 13th 2009, 04:09 AM

Hello, jzellt!

I "eyeballed" the solution . . .

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

Take the differences of consecutive terms.
Then take the differences of the differences ... and so on.

. . $\begin{array}{cccccccccc}\text{Sequence} & 5 && 10 && 17 && 26 && 37 \\ \text{1st diff.} & & 5 && 7 && 9 && 11 \\ \text{2nd diff.} & & & 2 & & 2 & & 2 \end{array}$

We see that the second differences are constant.
This indicates that the generating function is of the second degree, a quadratic.
. . That is, $f(n)$ contains $n^2.$

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

. . $\begin{array}{c|ccc} n & & a_n \\ \hline 1 & 5&=&2^2+1 \\ 2 & 10 &=&3^2+1 \\ 3 & 17 &=&4^2+1 \\ 4 & 26 &=&5^2+1 \\ 5 & 37 &=&6^2+1 \end{array}$
. . . $\begin{array}{c|c}\vdots & \vdots \\ n & \;\;(n+1)^2+1 \end{array}$

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

. . which verifies u2_wa's result.

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