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.

Note that starting with your first element, $a_0=1$; to get to $a_1=2$ we sum $1$, to get to $a_2=4$, we sum $2$, to get to $a_3=7$, we sum $3$. So in general, we can say that
$$a_{n}=a_{n-1}+n$$
This is
$$
\begin{align}
2&=1+1 \\[8pt]
4& =2+2\\[8pt]
7&=4+3\\[8pt]
11& =7+4\\[8pt]
\cdots&=\cdots\\[8pt]
a_n&=a_{n-1}+n
\end{align}
$$
We can get several solutions to you problem, I will share $2$:
Solution 1 Use the recursion to obtain a closed formula:
Since we know that $a_{n}=a_{n-1}+n$ we can write
$a_{n-1}=a_{n-2}+n-1$
so we get $$a_{n}=a_{n-2}+(n-1)+n$$
Repeating this process, we get that
$$a_{n}=a_{n-3}+(n-2)+(n-1)+n$$
$$a_{n}=a_{n-4}+(n-3)+(n-2)+(n-1)+n$$
$$a_{n}=a_{n-5}+(n-4)+(n-3)+(n-2)+(n-1)+n$$
...so in general we can say that
$$a_n=a_{n-k}+(n-k+1)+\cdots+n$$ for any $k$ a natural number.
(actually we should be proving the above by induction, but let it be)
So we can choose $k=n$, which means...
$$a_n=a_{n-n}+(n-n+1)+\cdots+n$$
$$a_n=a_0+(1+\cdots+n)$$
$$a_n=1+\frac{n(n+1)}{2}$$
Solution 2
Starting from $$a_{n}=a_{n-1}+n$$ we use generating functions:
$$\eqalign{
  & \sum\limits_{n = 1}^\infty  {{a_n}} {x^n} = \sum\limits_{n = 1}^\infty  {{a_{n - 1}}} {x^n} + \sum\limits_{n = 1}^\infty  n {x^n}  \cr 
  & \sum\limits_{n = 1}^\infty  {{a_n}} {x^n} = x\sum\limits_{n = 1}^\infty  {{a_{n - 1}}} {x^{n - 1}} + x\sum\limits_{n = 1}^\infty  {n{x^{n - 1}}}   \cr 
  & \sum\limits_{n = 0}^\infty  {{a_n}} {x^n} - {a_0} = x\sum\limits_{n = 0}^\infty  {{a_n}} {x^n} + x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty  {{x^n}}   \cr 
  & A\left( x \right) - 1 = xA\left( x \right) + x\frac{d}{{dx}}\frac{1}{{1 - x}}  \cr 
  & A\left( x \right) - xA\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}}  \cr 
  & \left( {1 - x} \right)A\left( x \right) = 1 + \frac{x}{{{{\left( {1 - x} \right)}^2}}}  \cr 
  & A\left( x \right) = \frac{1}{{1 - x}} + \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $$
The generating sequence of $f(x)=\frac{1}{{1 - x}}$ is $a_n=1$, while the generating sequence of $g(x)=\frac{x}{{{{\left( {1 - x} \right)}^3}}}$ can be obtained by diferentiation of the first one:
$$\eqalign{
  & \sum\limits_{n = 0}^\infty  {{x^n}}  = \frac{1}{{1 - x}}  \cr 
  & \sum\limits_{n = 1}^\infty  {n{x^{n - 1}}}  = \frac{1}{{{{\left( {1 - x} \right)}^2}}}  \cr 
  & \sum\limits_{n = 2}^\infty  {n\left( {n - 1} \right){x^{n - 2}}}  = \frac{2}{{{{\left( {1 - x} \right)}^3}}}  \cr 
  & \sum\limits_{n = 2}^\infty  {\frac{{n\left( {n - 1} \right)}}{2}{x^{n - 1}}}  = \frac{x}{{{{\left( {1 - x} \right)}^3}}}  \cr 
  & \sum\limits_{n = 1}^\infty  {\frac{{n\left( {n + 1} \right)}}{2}{x^n}}  = \frac{x}{{{{\left( {1 - x} \right)}^3}}} \cr} $$
so we finally get that
$$a_n=1+{\frac{{n\left( {n + 1} \right)}}{2}}$$
ADD I forgot to add the sum of the $a_n$s!
You need to evaluate
$$\sum_{k=0}^{n-1} a_k=\sum_{k=0}^{n-1} 1+\sum_{k=0}^{n-1}\frac{k(k+1)}{2}$$
$$\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}\frac{k(k-1)}{2}$$
$$\sum_{k=0}^{n-1} a_k=n+\sum_{k=0}^{n}{k\choose 2}$$
Using the binomial identity
$$\sum_{k=0}^n {k\choose l}={{n+1}\choose {l+1}}$$ we get
(you can find how to obtain it here
$$\sum_{k=0}^{n-1} a_k=n+{n+1\choose 3}$$
$$\sum_{k=0}^{n-1} a_k=n+\frac{(n+1)n(n-1)}{6}$$
which is what you wanted.
