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.

(Edit: Upps, I see now this is essentially solution (2) of Peter Tamaroff's answer, but because it's much shorter I just leave it here) 
Your sequence can be separated into 2 sequences, where we add each pair:
$\begin{eqnarray}
 &1&2&4&7&11&16&\cdots & = a_k\\
\hline 
=&1&1&1&1&1&1&\cdots \\
+&0&1&3&6&10&15&\cdots \\
\hline \end{eqnarray}$    
Then the partial sums are, beginning the index k at 1:
$\begin{eqnarray}
 &1&3&7&14&25&41&\cdots &=&s_k\\
\hline 
=&1&2&3&4&5&6&\cdots &= &&=&k\\
+&0&1&4&10&20&35&\cdots &=&\binom{1+k}{3}&=&{(k+1)!\over 3! (k-2)!}\\
\hline 
=&1&3&7&14&25&41&\cdots &=&s_k&=& k+ {(k+1)!\over 3! (k-2)!}\\
\end{eqnarray}$        
The last formula can be simplified to 
$$ s_k = k+ {(k+1)k(k-1) \over 6} = {6k+k^3-k\over 6} 
 = k \cdot {k^2+5\over 6}$$
