Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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?

share|improve this question
2  
Hint: take the difference of succeeding terms. –  Thomas Andrews Dec 4 '11 at 15:34
2  
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
4  
The answer is 42. For every $n$. –  Did Dec 4 '11 at 15:56
4  
The OEIS is great for questions like these... –  J. M. Dec 4 '11 at 16:07
1  
Pedants are NOT great for questions like these... –  The Chaz 2.0 Dec 4 '11 at 21:03
show 1 more comment

3 Answers

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

share|improve this answer
    
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
add comment

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 $a_n=-3+{(n+2)(n+3)\over2}$.

share|improve this answer
add comment

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

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.