How do I find the sum of a sequence whose common difference is in Arithmetic Progression ?
Like in the following series :-
$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91$
And also how to find it's $n^{th}$ term ??
[Math] How to find the sum of a sequence whose common difference is in Arithmetic Progression
sequences-and-series
Best Answer
The sequence that you are talking about is a quadratic sequence. A quadratic sequence is a sequence of numbers in which the second difference between any two consecutive terms is constant (definition taken from here).
The difference of consecutive terms in your sequence forms an arithmetic progression $2,3,4,5,\dots$ with common difference of $1$. Since the sequence is a quadratic sequence, the $n^{th}$ term of the sequence is given by a quadratic polynomial: $$T_n = an^2 + bn + c$$ As stated by Siddhant in his answer, you could just plug in $n$ as $1,2$ and $3$ to get three equations in three variables and get the values of $a,b$ and $c$. However, we could use a rather generalized formula for getting the equation for $n^{th}$ term of the sequence (you should try deriving this formula): $$T_n = \left(\frac{d_0}{2}\right)n^2 + \left(d - \frac{3\cdot d_0}{2}\right)n + (a + d_0 - d)$$ where $a$ is the first term of the sequence, $d = T_2 - T_1$ i.e. difference between first two terms of the sequence and $d_0$ is the second difference between any two consecutive terms of the sequence. In your case, $a = 1$, $d = 2$ and $d_0 = 1$. Plugging in these values in the equation yields $$T_n = \frac{1}{2}n^2 + \frac{1}{2}n$$
For finding the sum: $$\sum_{i = 1}^{n}T_i$$ $$=\sum_{i = 1}^{n}\left(\frac{1}{2}i^2 + \frac{1}{2}i\right)$$ You can solve this using the already known summations $\sum_{i = 1}^ni^2$ and $\sum_{i = 1}^ni$.