# What is the formula for the nth term of the sequence 1, 3, 6, 10, 15,... ?

Jan 25, 2017

${a}_{n} = \frac{1}{2} n \left(n + 1\right)$

#### Explanation:

These are triangular numbers:

$0 \textcolor{w h i t e}{0000} 0 \textcolor{w h i t e}{000000} 0 \textcolor{w h i t e}{00000000} 0 \textcolor{w h i t e}{0000000000} 0$
$\textcolor{w h i t e}{0000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{000000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{00000000} 0 \textcolor{w h i t e}{0} 0$
$\textcolor{w h i t e}{0000000000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{000000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0$
$\textcolor{w h i t e}{000000000000000000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0$
$\textcolor{w h i t e}{0000000000000000000000000000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0$

Geometrically, you can see that such a triangle is one half of a parallelogram with base $n + 1$ and height $n$, e.g. ${a}_{5} :$

$\textcolor{w h i t e}{0000} 0 \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0}$
$\textcolor{w h i t e}{000} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0}$
$\textcolor{w h i t e}{00} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0}$
$\textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} \textcolor{b l u e}{0} \textcolor{w h i t e}{0} \textcolor{b l u e}{0}$
$0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} 0 \textcolor{w h i t e}{0} \textcolor{b l u e}{0}$

Such a parallelogram has a total count of $n \left(n + 1\right)$, so the triangle is half of that.

In other words:

${a}_{n} = {\sum}_{k = 1}^{n} k = \frac{1}{2} n \left(n + 1\right)$

Jan 25, 2017

$\frac{{n}^{2} + n}{2}$

#### Explanation:

Starting with 1 you can begin to see a pattern develop....

${a}_{1} = 1$

${a}_{2} = {a}_{1} + 2 = 1 + 2 = 3$

${a}_{3} = {a}_{2} + 3 = 3 + 3 = 6$

${a}_{4} = {a}_{3} + 4 = 6 + 4 = 10$

${a}_{5} = {a}_{4} + 5 = 10 + 5 = 15$

$\vdots$

${a}_{n} = {a}_{n - 1} + n$

Since every term is merely the sum of the previous term plus the next counting number we have.

${a}_{n} = {\sum}_{k = 1}^{n} k = \frac{n \left(n + 1\right)}{2}$

To test this we use the first formula for ${a}_{n + 1}$

${a}_{n + 1} = {a}_{n} + \left(n + 1\right)$

$\implies {a}_{n + 1} = \frac{n \left(n + 1\right)}{2} + \left(n + 1\right) = \frac{{n}^{2} + n}{2} + \frac{2 n + 2}{2} = \frac{{n}^{2} + 3 n + 2}{2}$

by the hypothesis

${a}_{n + 1} = \frac{\left(n + 1\right) \left(n + 2\right)}{2} = \frac{{n}^{2} + 3 n + 2}{2}$

They are equal...

So the hypothesis is proven

This is an example of "Weak" Induction