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

Question

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

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?

Answer 1

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

Answer 2

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

Answer 3

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

$t_0=0$

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

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

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

...

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

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

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

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

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

Answer 4

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

Answer 5

$ 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,$$\sum_{n=0}^{10} a_n = \frac{n^2 + 5n}{2} $$