$$3, 7, 12, 18, 25, \ldots$$
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?
$\endgroup$ 65 Answers
$\begingroup$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 $n=0,1,2....$
$\endgroup$ 2 $\begingroup$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 $a_n=-3+{(n+2)(n+3)\over2}$.
$\endgroup$ $\begingroup$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 $$x_n + 3 = 1 + 2 + x_n = \underbrace{1+2+3+\dots}_{n+2 \text{ terms}}$$
$\endgroup$ 2 $\begingroup$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
$t_0=0$
$t_1=t_0+(1+2)=(1+2)$
$t_2=t_1+(2+2)=(1+2)+(2+2)$
$t_3=t_2+(3+2)=(1+2)+(2+2)+(3+2)$
...
$t_n=t_{n-1}+(n+2)$
$=(\color{red}1+\color{green}2)+(\color{red}2+\color{green}2)+(\color{red}3+\color{green}2)+...+(\color{red}n+\color{green}2)$
$=\color{red}{(1+...+n)}+\color{green}{(2n)}$
$=\frac{n(n+1)}{2} + 2n$
$=\frac{(n^2+5n)}{2}$
$\endgroup$ $\begingroup$$ S = 3 + 7 + 12 + 18+ \cdots n^{th}$ term
$S = 0 + 3 + 7 + 12 + 18 + \cdots n^{th} $ term
By subtracting the above to equation we get,
$a_n = 3 + 4 + 5 + \cdots n\, terms$ (Here, $a_n$ is the general term)
$a_n = \frac{(n)(5+n)}{2} $
Now,$$\sum_{n=0}^{10} a_n = \frac{n^2 + 5n}{2} $$
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