# Thread: How to form the correct equation

1. ## How to form the correct equation

I suspect I am in the correct forum. However, due to partial or complete ignorance on my part, of disciplines above and including algebra, I may not be.
So in an effort save some time for this particular situation, I am presenting this here, before I embark on the learning curve,...if that is even possible.

Trying to define the equation I would use to find the value of a given number, using the following logic:

Given Number: 2 its value is: 1 determined via 1
Given Number: 3 its value is: 3 determined via 1+2
Given Number: 4 its value is: 6 determined via 1+2+3
Given Number: 5 its value is: 10 determined via 1+2+3+4
... to infinity

This feels obvious,...but I just can't seem to see it.

2. Try $\displaystyle t_n= 1+2+3+\cdots +(n-1)+ n = \frac{n(n+1)}{2}$

3. Originally Posted by pickslides
Try $\displaystyle t_n= 1+2+3+\cdots +(n-1)+ n = \frac{n(n+1)}{2}$
$\displaystyle \frac{n(n+1)}{2}{-n}$ produces the correct result.

However, (and again please excuse my ignorance), it is unclear to me how you arrived at: $\displaystyle \frac{n(n+1)}{2}$

from: $\displaystyle 1+2+3+\cdots +(n-1)+ n$

4. Originally Posted by spfaff
[tex]However, (and again please excuse my ignorance), it is unclear to me how you arrived at: $\displaystyle \frac{n(n+1)}{2}$
from: $\displaystyle 1+2+3+\cdots +(n-1)+ n$
That is just one of the best known formulas in all mathematics.
It in easily proven using mathematical induction.

5. There is a story that when Gause was 8 years old, just to keep the students busy, his teacher assigned them the problem of adding all integer from 1 to 100. Gause immediately wrote 5050 on his slate, turned it over, and sat there. Presumably, he apparently recognized that if he reversed the sum he would have
$\begin{array}{ccccc}1+ & 2+ & + 3& +\cdot\cdot\cdot &+ 100 \\ 100+ & 99+ & 98+ \cdot\cdot\cdot&+ 1\\ 101 & + 101& + 101& \cdot\cdot\cdot & 101\end{array}$
That is, each pair adds to 100 and there are 100 such pairs: 101(100)= 10100 and since we have added twice each sum is 10100/2= 5050.

Similarly, you can continue to "n" rather than 100 to get
$\begin{array}{ccccc}1+ & 2+ & + 3& +\cdot\cdot\cdot &+ n \\ n+ & n-1+ & n-2+ \cdot\cdot\cdot&+ 1\\ n+1 & + n+1&+ n+ 1&+ n+1& \cdot\cdot\cdot & n+1\end{array}$
so there are n pairs, each summing to n+1: the total of the pairs is n(n+1) and the sum of each is
$\frac{n(n+2)}{2}$