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$\begingroup$

$3, 7, 12, 18, 25, \ldots$

This sequence appears in my son's math homework. The question is to find the $n$'th term. What is the formula and how do you derive it?

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    Pedants are NOT great for questions like these...2011-12-04

5 Answers 5

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Recursive formula is given by following expression .

$a_n=(n+3)+a_{n-1}$ ; with $a_0=3$

EDIT :

According to WolframAlpha closed form is :

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

where $n=0,1,2....$

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    The $n$th term is $a_n$.2011-12-04
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If you stare hard at the sequence long enough, you'll realize it is $ \underbrace{3}_{a_3},\underbrace{(3+4)}_{a_4},(3+4+5),(3+4+5+6),\ldots ,\underbrace{(3+4+5+\cdots+n)}_{a_n} $ (I start counting at 3 for clarity)

So, $\tag{1}a_n=3+(4+5+\cdots+n)=-3+ (1+2+3+\cdots+n).$

Now suppose $n$ is even. Then we can group the numbers in the sum $1+2+\cdots+n$ as follows: $\color{green}1+\color{red}2+\color{blue}3+\color{pink}4+\color{orange}5+\cdots +\color{orange}{(n-4)}+\color{pink}{(n-3)}+\color{blue}{(n-2)}+\color{red}{(n-1)}+\color{green}n$

The sum of each group of the same color is $n+1$ and there are $n\over2$ groups. So, $ 1+2+3+\cdots+n={n(n+1)\over 2}, \text{ for }n \text{ even.} $

For $n$ odd, $\eqalign{ 1+2+3+\cdots +n&= \bigl[ 1+2+3+\cdots(n-1)\Bigr]+n\cr &= {(n-1)\bigl((n-1)+1\bigr)\over2}+n\cr &={n(n+1)\over2},}$ where we used the result for the even case in the second line.

Combining this result with (1): $ a_n=-3+{n(n+1)\over 2}, $ where $a_3$ is the first term.

If you want the first term of the sequence to be $a_1$, then $a_n=-3+{(n+2)(n+3)\over2}$.

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Do you know the closed form for the triangular numbers? This sequence is three less than the $n+2$ triangular number.

Your sequence can be written: $3,3+4,3+4+5,3+4+5+6,3+4+5+6+7,\dots$

Then general $n$th term is:

$x_n = \underbrace{3+4+5...}_{n \text{ terms}}$

So $x_n + 3 = 1 + 2 + x_n = \underbrace{1+2+3+\dots}_{n+2 \text{ terms}}$

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    @Veronica To find out more about triangular numbers you can have a look at this question: http://math.stackexchange.com/questions/60578/what-is-the-term-for-a-factorial-type-operation-but-with-summation-instead-of-p and this question: http://math.stackexchange.com/questions/78936/finding-a-formula-to-sum-natural-numbers-up-to-n2011-12-05
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for the series: 3, 7, 12, 18, 25, …

$3=0+3\\7=3+4\\12=7+5\\18=12+6\\25=18+7$

We can find that: each term equals previous term plus n+2

$t_0=0$

$t_1=t_0+(1+2)=(1+2)$

$t_2=t_1+(2+2)=(1+2)+(2+2)$

$t_3=t_2+(3+2)=(1+2)+(2+2)+(3+2)$

...

$t_n=t_{n-1}+(n+2)$

$=(\color{red}1+\color{green}2)+(\color{red}2+\color{green}2)+(\color{red}3+\color{green}2)+...+(\color{red}n+\color{green}2)$

$=\color{red}{(1+...+n)}+\color{green}{(2n)}$

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

$=\frac{(n^2+5n)}{2}$