How do I find the sum of a sequence whose common difference is in Arithmetic Progression? 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 ??
 A: 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 $\sum_{i = 1}^ni$.
A: You might be interested to know that if the differences of consecutive terms of a sequence are given by some polynomial, i.e $a_{n+1}-a_n=P(n)$, for some polynomial $P$, then the terms of the sequence themselves are given by a polynomial, with one greater degree. As in your case, the differences of terms are in A.P, i.e the differences are given by a linear polynomial: 
$$a_{n+1}-a_n=P(n)=n+1$$
So, the general term of the sequence will be given by a quadratic polynomial $an^2+bn+c$. All you have to do is find the constants $a,b,c$ which you can do using the first three terms of the sequence. As for summation of this sequence, use the known summations $\sum_{k=1}^nk$ and $\sum_{k=1}^nk^2$.
A: In such a case, every generic term of the sequence can be represented as
$An^2 + Bn + C$
where A B C are fixed values and n is the position of the term. If the difference of difference of the terms were in an ap, the generic term would be
$An^3 + Bn^2 + Cn + D$
Where A B C and D are fixed values.
This will keep on going till the differences are in AP.
A: I enjoyed several of the solutions here. However the idea of substituting values and solving for a, b, c seems really long and inefficient.
Below I have an efficient method to do this. 
As codetalker pointed out:
a(n+1)-a(n)= some linear term=1+n(for example) 
So now what you do is make a telescopic series:
a(2)-a(1)=1+1
a(3)-a(2)=1+2
........................
a(n)-a(n-1)=1+(n-1)
So now you add all individual equations, you see that
a(n) =a(1)+n×1+(n-1)(n)/2 
( I have used the formula for sum of n terms).
You know a(1)
So now you have a(n) 
On summation you can apply the formulae the other answers specified for sum of numbers and squares.
Hope this answer makes is more efficient to do these
A: $$
\begin{align}
& \text{Let } a = a_{0} \space\,,\space d = \frac{1}{n} (a_{n} - a_{n-1}) \space\Rightarrow\space a_{n} = a_{n-1} + d \cdot n \quad \colon n \ge 1 \\
& \small \color{blue}{ (a=0 \,, d=1) \Rightarrow a_{n} = a_{n-1} + n = \{ \text{1, 3, 6, 10, ...} \} } \\
& \\
& a_{n} = d (n) + a_{n-1} = d (n) + d (n-1) + a_{n-2} = d (n) + d (n-1) + \text{...} + d (2) + d (1) + a \Rightarrow \\
& a_{n} = a + d \sum_{k=1}^{n} (k) = a + \frac{d}{2} n (n+1) \\
& \small \color{blue}{ (a=0 \,, d=1) \Rightarrow a_{n} = \frac{1}{2} n(n+1) = \{ \text{1, 3, 6, 10, ...} \} } \\
& \\
& \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} \left( a + \frac{d}{2} k(k+1) \right) = a \cdot n + \frac{d}{2} \left[ \sum_{k=1}^{n} k + \sum_{k=1}^{n} k^2 \right] \Rightarrow \\
& \sum_{k=1}^{n} a_{k} = a n + \frac{d}{2} \left[ \frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6} \right] = a n + \frac{d}{6} n(n+1)(n+2) \\
& \small \color{blue}{ (a=0 \,, d=1) \Rightarrow \sum_{k=1}^{n} a_{k} = \frac{1}{6} n(n+1)(n+2) = \{ \text{1, 4, 10, 20, ...} \} }
\end{align}
$$
A: Empirically:
By educated guess, as the differences grows as $n$, the terms grow as $n^2$. Taking the ratios $T_n/n^2$, you see them converging to $1/2$.
Then the sequence of $2T_n-n^2$ is
$$1,2,3,4,5,6,7,\cdots$$
