Doubt about series - which series is this? 
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Sum of  n consecutive numbers 

I really can't remember (if I have ever known this): which series is this and how to demonstrate its solution?
$$\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$$
 A: Write out this sum twice, once is direct order, and once in reverse:
$$
   \begin{align}
       &1 + &2 + &\ldots+ &(n-1)+ &n &=s \\
       &n + &(n-1)+ & \ldots+ &2+ &1 &=s 
   \end{align}
$$
Now add up column-wise:
$$
    (n+1) + (n+1) + \ldots + (n+1) + (n+1) = 2s
$$
There are exactly $n$ terms here (as many as the number of terms in the sum). Hence:
$$
    n(n+1) = 2s
$$
Now solve for $s$.
A: This is an Arithmetic Series starting from $1$ with difference $1$.
$$ \sum\limits_{i=1}^n i = 1 + 2 + 3 + 4 + ... +n = {n (n+1) \over 2}$$
1#Check this out.
2#There's one another related one.
A: This is not a series. This sum is named Gauss sum and that formula $\frac{n(n+1)}{2}$ you can prove it using induction. 
The exercise starts from the following sum: $1+2+ \ldots +100$ and the way you can classify the terms of this sum. 
$1+2+ \ldots + 100 = (1+100)+ (2+99)+ \ldots (50+51$). 
for more information I think the following link : http://www.newton.dep.anl.gov/askasci/math99/math99155.htm it is a good one. 
A: $S = 1 + 2 + 3 + \ldots + n = n + (n-1) + (n-2) + \ldots + 1$.  So $2S = (n+1) + \ldots + (n+1)$.  Since the $(n+1)$ appears $n$ times, $$S = \frac{n(n+1)}{2}$$
