# How do you find the explicit formula for the following sequence 0, 1, 3, 6, 10 ...?

Feb 7, 2016

For $n = 1$, ${a}_{1} = 0$
For $n > 1$, ${a}_{n} = {a}_{1} + \sum x$ where $x \in \mathbb{N}$ and $x < n$

#### Explanation:

Express the desired element in terms of the preceding element

${a}_{1} = 0$

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

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

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

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

Based from the pattern, we can say that

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

If we keep on expanding the formula, we have

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

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

$\implies {a}_{n} = \left(\left(\left({a}_{n - 4} + \left(n - 4\right)\right) + \left(n - 3\right)\right) + \left(n - 2\right)\right) + \left(n - 1\right)$

Eventually we will have

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

Feb 14, 2016

Examine sequences of differences to derive an explicit formula:

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

#### Explanation:

A finite sequence does not determine a unique formula, but in this particular case there are enough terms to see an intended pattern.

We can derive an explicit formula by examining sequences of differences.

The given sequence is:

$\textcolor{b l u e}{0} , 1 , 3 , 6 , 10$

Write down the sequence of differences of that sequence:

$\textcolor{b l u e}{1} , 2 , 3 , 4$

Write down the sequence of differences of the sequence:

$\textcolor{b l u e}{1} , 1 , 1$

Having arrived at a constant sequence, we can now use the initial term of each of these sequences as coefficients to construct an explicit polynomial formula:

a_n = color(blue)(0)/(0!) + color(blue)(1)/(1!)(n-1) + color(blue)(1)/(2!)(n-1)(n-2)

$= n - 1 + \frac{1}{2} {n}^{2} - \frac{3}{2} n + 1$

$= \frac{1}{2} {n}^{2} - \frac{1}{2} n$

$= \frac{1}{2} n \left(n - 1\right)$