Sum of terms whose differences are in Arithmetic progression.

Question

My friend gave me a problem of finding sum of 10 terms. I simplified the terms and got:
Terms: 3, 9, 18, 30, ... , 165
Differences: 6, 9, 12, ... , 30

I saw that the terms' differences are in AP. I got the sum 660 by simply adding them. Then I took some small numbers whose differences were in AP and tried to make a formula to find their sum. I got the following formula:

Sum of n terms (T_n) = $n$a₁ $ + \sum_{i=1}^{n-1} d_i + \sum_{i=1}^{n-2} d_i + \sum_{i=1}^{n-3} d_i + ...$$+ \sum_{i=1}^1 d_i$
Where d_n is term of differences. Means, the terms of 6, 9, 12, ... , 30.
And a₁ is first term of given numbers (Here -> 3).

I tried it in different examples and it works. But this can't work if there are a hunded terms. I want to make the formula more smaller and accurate. What should I do?

Answer 1

If the 2nd difference is constant, we have a quadratic sequence. Meaning, the $n$th term is $an^2 + bn + c$ for some $a,b,c$.

The 2nd difference here is 3, and the $n^2$ coefficient is always half the 2nd difference. So, $a=\frac{3}{2}$.

If we write the sequence $\frac{3}{2}n^2$, we get: $1.5, 6, 13.5, 24,\ldots$ and if we take the original sequence, and subtract this sequence we have the sequence $bn+c$, with terms $1.5, 3, 4.5, 6, \ldots$. This is clearly a linear sequence, with common difference $1.5$. So in fact, $b=1.5, c=0$.

Therefore, the $n$th term of your sequence is $$\frac{3n(n+1)}{2}.$$

Writing it like this, you may notice that this is just the triangle numbers multiplied by $3$, since the $n$th triangle number is $\frac{n(n+1)}{2}$. Can you use this information to find a formula for the sum of $n$ terms?