Sequences - nth Term

**kate** · June 10th 2007, 01:47 AM

Hi. Please can you help me? I have three Maths tests next week, and we have a revision sheets which says lots of stuff, and the last one I can't do! It says nth term. I can do nth term all right, but I was wondering if there is an equation for a sequence if each time the gap changes, eg;
8, 11, 15, 20, 26
+3 +4 +5 +6

__n +/- ___.

Thanks. K

**CaptainBlack** · June 10th 2007, 03:48 AM

Originally Posted by kate

Hi. Please can you help me? I have three Maths tests next week, and we have a revision sheets which says lots of stuff, and the last one I can't do! It says nth term. I can do nth term all right, but I was wondering if there is an equation for a sequence if each time the gap changes, eg;
8, 11, 15, 20, 26
+3 +4 +5 +6

__n +/- ___.

Thanks. K

$a_1=8$

$a_2=8+(2+1)$

$a_3=8+(2+1)+(2+2)=8+2\times2+(1+2)$

$a_4=8+(2+1)+(2+2)+(2+3)=8+3\times2+(1+2+3)$

:
:

$a_n=8+(n-1)\times 2+\sum_{r=1}^{n-1}r$

Where the last term is the sum of the first $n-1$ integers so:

$a_n = 8+2(n-1)+\frac{(n-1)n}{2}$

RonL

**Soroban** · June 10th 2007, 06:36 AM

Hello, Kate!

Here's another approach . . .

Take the difference of consecutive terms.
Then take the differences of the differences, and so on
. . until you get a row of constants.

. . $\begin{array}{cccccccccc}\text{Sequence:} & 8 & & 11 & & 15& & 20 & & 26 \\<br /> \text{1st di{f}ferece:} & & 3 & & 4 & & 5 & & 6 &\\<br /> \text{2nd dif{f}erence:} & & & 1 & & 1 & & 1 & & \end{array}$

The row of constants appeared in the second differences.
. . Hence, the generating function is of the second degree, a quadratic.

There are a number of methods for determining the generating function.
. . Here's one of them . . .

We know that the function has the form: . $f(n) \:=\:an^2 + bn + c$

Using the first three values of the sequence, we have:
. . $\begin{array}{ccccccc}f(1) = 8: & a + b + c & = & 8 & [1]\\<br /> f(2) = 11: & 4a + 2b + c & = & 11 & [2]\\<br /> f(3) = 15: & 9a + 3b + c & = & 15& [3]\end{array}$

Then solve the system of equations . . .

Subtract [1] from [2]: . $3a + b \:=\:3\;\;\;[4]$
Subtract [2] from [3]: . $5a + b \:=\:4\;\;\;[5]$

Subtract [4] from [5]: . $2a = 1\quad\Rightarrow\quad\boxed{a = \frac{1}{2}}$

Substitute into [4]: . $3\left(\frac{1}{2}\right) + b\:=\:3\quad\Rightarrow\quad\boxed{b = \frac{3}{2}}$

Substitute into [1]: . $\frac{1}{2} + \frac{3}{2} + c \:=\:8\quad\Rightarrow\quad\boxed{c = 6}$

Therefore: . $f(n) \;=\;\frac{1}{2}n^2 + \frac{3}{2}n + 6\quad\Rightarrow\quad \boxed{f(n)\;=\;\frac{1}{2}(n^2 + 3n + 12)}$

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