- #1

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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

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...