Solve the problem involving sum of a series

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Homework Statement
Simplify ##r(r+1)-(r-1)r## and use your result to obtain ## \sum_{r=1}^n r##
Relevant Equations
Method of difference
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My attempt;

##r^2+r-r^2+r=2r##

Let ##f(r)=(r-1)r## then it follows that ##f(r+1)=r(r+1)## so that ##2r## is of the form ##f(r+1)-f(r)##.

When

##r=1;## ##[2×1]=2-0##
##r=2;## ##[2×2]=6-2##
##r=3;## ##[2×3]=12-6##
##r=4;## ##[2×4]=20-12##

...

##r=n-1##, We shall have ##2(n-1)=n-1(n)-(n-2)(n-1)##

##r=n##, We shall have ##2n=n(n+1)-(n-1)n## ## 2\sum_{r=1}^n r=n(n+1)-0##

## 2\sum_{r=1}^n r=n^2+n##

## \sum_{r=1}^n r=\dfrac{1}{2} \left[n^2+n\right]##

your insight is welcome...
 
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I mean, do we have to make things complicated here or what?
 
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chwala said:
Homework Statement:: Simplify ##r(r+1)-(r-1)r## and use your result to obtain ## \sum_{r=1}^n r##
Relevant Equations:: Method of difference

Let f(r)=(r−1)r then it follows that f(r+1)=r(r+1) so that 2r is of the form f(r+1)−f(r).
2r=f(r+1)-f(r)
2\sum_{r=1}^nr=\sum_{r=1}^nf(r+1)-\sum_{r=1}^nf(r)=\sum_{r=1}^nf(r+1)-\sum_{r=0}^{n-1}f(r+1)=f(n+1)-f(1)=n(n+1)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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