Sum of the series : $1 + 2+ 4 + 7 + 11 +\cdots$ I got a question which says
$$ 1 + \frac {2}{7} + \frac{4}{7^2} + \frac{7}{7^3} + \frac{11}{7^4} + \cdots$$
I got the solution by dividing by $7$ and subtracting it from original sum. Repeated for two times.(Suggest me if any other better way of doing this). 
However now i am interested in understanding the series 1,2,4,7,11,..... In which the difference of the numbers are consecutive natural numbers. 
How to find the sum of 
$1+2+4+7+11+\cdots nterms$
This is my first question in MSE. If there are some guidelines i need to follow, which i am not, please let me know.

Let's first find the general form of the terms of the sum 
$1 + 2+ 4 + 7 + 11 +\ldots.$
The terms obey the recurrence
$a_n = a_{n-1} + n$, where $a_0 = 1.$
Using standard techniques we find 
$$a_n = \frac{1}{2}n(n+1) + 1.$$
These are basically the triangular numbers, as indicated by @anon in the comments.
The sum of the first $n+1$ terms can be found using 
Faulhaber's formula, as indicated by @RossMillikan, 
$$\begin{eqnarray*}
\sum_{k=0}^n a_k 
&=& \frac{1}{2}\sum_{k=0}^n n^2 
    + \frac{1}{2}\sum_{k=0}^n n
    + \sum_{k=0}^n 1 \\
&=& \frac{1}{2}\frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}\frac{1}{2} n(n+1) + n+1 \\
&=& \frac{1}{6}(n+1)(n^2+2n+6).
\end{eqnarray*}$$
The sum
$$\begin{equation*}
\sum_{n=0}^\infty \frac{a_n}{7^n}
= \frac{1}{2}\sum_{n=0}^\infty \frac{n^2}{7^n}
    +  \frac{1}{2}\sum_{n=0}^\infty \frac{n}{7^n}
    + \sum_{n=0}^\infty \frac{1}{7^n} \tag{1}
\end{equation*}$$
can be found by a standard trick.
Consider the geometric series
$$\sum_{k=0}^\infty x^n = \frac{1}{1-x}$$
for $|x|<1$.
The last sum on the right hand side of (1) is just $\displaystyle\frac{1}{1-\frac{1}{7}} = 7/6$.
Now notice that for $m\in \mathbb{N}$
$$\begin{eqnarray*}
\sum_{n=0}^\infty n^m x^n
&=& \left(x \frac{d}{dx}\right)^m \sum_{k=0}^\infty x^n \\
&=& \left(x \frac{d}{dx}\right)^m \frac{1}{1-x}.
\end{eqnarray*}$$
Therefore,
$$\begin{eqnarray*}
\sum_{n=0}^\infty \frac{a_n}{7^n}
&=& \left[\frac{1}{2} \left(x \frac{d}{dx}\right)^2 \frac{1}{1-x}
+ \frac{1}{2} \left(x \frac{d}{dx}\right) \frac{1}{1-x}
+ \frac{1}{1-x} \right]_{x=1/7} \\
&=& \frac{1}{2}\frac{7}{27} + \frac{1}{2} \frac{7}{36} + \frac{7}{6} \\
&=& \frac{301}{216}.
\end{eqnarray*}$$
