# How to find $n$'th term of the sequence $3, 7, 12, 18, 25, \ldots$?

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

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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Hint: take the difference of succeeding terms. – Thomas Andrews Dec 4 '11 at 15:34
In principle, there are infinitely many sequences that start like this, so you cannot "derive it", the task is to find a simple recipe that gives this sequence. There is little point in looking up the recipe, but I am sure someone will provide it. – Phira Dec 4 '11 at 15:35
The answer is 42. For every $n$. – Did Dec 4 '11 at 15:56
The OEIS is great for questions like these... – J. M. Dec 4 '11 at 16:07
Pedants are NOT great for questions like these... – The Chaz 2.0 Dec 4 '11 at 21:03

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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And the $nth$ term? – Veronica Dec 4 '11 at 15:50
The $n$th term is $a_n$. – The Chaz 2.0 Dec 4 '11 at 21:03

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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Wolfram Alpha recommends $\frac{1}{2}(n^2+5n)$ – Veronica Dec 4 '11 at 16:07
@Veronica To find out more about triangular numbers you can have a look at this question: math.stackexchange.com/questions/60578/… and this question: math.stackexchange.com/questions/78936/… – Martin Sleziak Dec 5 '11 at 12:36