Possible Duplicate:
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?
$$\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$$
Possible Duplicate:
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?
$$\sum\limits_{i=1}^n i = \frac{n(n+1)}{2}$$
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 $s$.
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.
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.
$S = 1 + 2 + 3 + \ldots + n = n + (n-1) + (n-2) + \ldots + 1$. So $2S = (n+1) + \ldots + (n+1)$. Since the $(n+1)$ appears $n$ times, $$S = \frac{n(n+1)}{2}$$