# Generating a closed form solution of triangular numbers

The Triangular numbers are known to be $1, 3, 6, 10, 15, 21 ...$ In general, the $n$th triangular number is $$\frac{n(n + 1)}{2}.$$ Is there a way to get the general closed form solution from a given seed value ?

For example, in the above numbers, the pattern is easy to see as $+2, +3, +4, +5, +6,\dotsc$ and so on.

What if I have some arbitrary starting value $A$ and a seed value such as $B$ so I can generate the following sequence $A, A + B, (A + B) + (B + 1)...$

What would be the closed form solution to this sequence similar to something like the one for triangular numbers?

• Search for "arithmetic progression" Oct 19 '17 at 20:29
• @Chappers In an arithmetic progression the difference between any two given terms in constant. That is not the case here. Oct 19 '17 at 20:31
• Whoops, good point. So what you have is actually the sum of the first $n$ terms of an arithmetic progression. Oct 19 '17 at 20:34
• You can use the same proof as for the triangular number formula: write them out in both original and reverse order, add the corresponding terms, and divide by 2. Oct 19 '17 at 20:36

You can take out A to see the sequence more clearly:

$0,B,B+B+1,B+B+1+B+2,...$

So, at the $n-th$ step (starting from $0$-th step) we have the term $nB+(1+2+...+(n-1))$

Hence, overall the $n-th$ term is $A+ nB+(1+2+...+(n-1))$, which I'm sure you can simplify it further.

We can use the sum function of a general quadratic sequence:

$$S_2(n,s,d_1,c) = \cfrac{n(cn^2 + 2c + 3nd_1 + 6s - 3cn - 3d_1)}{6}$$

Where $n$ is the number of terms to be summed, $s$ is the starting term of the series, $d_1$ is the first difference (subtracting the first term from the second term) and $c$ is the constant difference between the differences.

So e.g. for $1, 3, 6, 10, 15, 21, \ ...$ you would get the $n$th number by taking the sum up to $n$ terms, then subtracting the sum up to $n-1$ terms.

For your example we can see that $s = 1, d_1 = 2, c=1$, plug in the values into the function:

$$S_2(n,1,2,1) - S_2(n-1,1,2,1) = \frac{n(n+1)}{2}$$

Here you can use any starting value, any difference and so on.