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.
 A: 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,
$$\begin{eqnarray*}
\sum_{n=0}^\infty \frac{a_n}{7^n}
&=& \left[\frac{1}{2} \left(x \frac{d}{dx}\right)^2 \frac{1}{1-x}
+ \frac{1}{2} \left(x \frac{d}{dx}\right) \frac{1}{1-x}
+ \frac{1}{1-x} \right]_{x=1/7} \\
&=& \frac{1}{2}\frac{7}{27} + \frac{1}{2} \frac{7}{36} + \frac{7}{6} \\
&=& \frac{301}{216}.
\end{eqnarray*}$$
A: Hint: your series is simply $\sum\limits_n^{k} \frac{n^2}{2}-\frac{n}{2}+1$
A: 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.
A: Your series (without the denominators) is $$\sum_{i=1}^n 1+ \frac {i(i-1)}2=\sum_{i=1}^n 1+\frac {i^2}2 - \frac i2=n+\frac {n(n+1)(2n+1)}{12}-\frac{n(n+1)}4$$  This is an application of Faulhaber's formula
A: Let T = 1 + 2/7 + 4/7^2 + 7/7^3 + ... ----> (1)
Then 
  T/7 =     1/7 + 2/7^2 + 4/7^3 + ... ----> (2)
Now, 
(2)-(1) => 6T/7 = 1 + 1/7 + 2/7^2 + 3/7^3 ----> (3)
Hence, (1/7)(6T/7) =  1/7 + 1/7^2 + 2/7^3 ----> (4)
So, (3)-(4)=> (1 - 1/7)(6T/7) = 1 + 0 + 1/7^2 + 1/7^3 + ...
       => (6/7)(6T/7) = 1 + (1/7^2 + 1/7^3 + .... infinity)

       => 36T/49 = 1 + (1/7^2)/(1 - 1/7)    [Formula = a/(1-r)]

       => 36T/49 = 1 + (1/49)/(6/7)

       => T = (49/36)(1 + (1/49)(7/6))

       => T = 49/36 + (49/36)(1/49)(7/6)

       => T = 49/36 + 7/216

       => T = (294 + 7)/216

       => T = 301/216 

A: (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 
$$ s_k = k+ {(k+1)k(k-1) \over 6} = {6k+k^3-k\over 6} 
 = k \cdot {k^2+5\over 6}$$
