# Solve the problem involving sum of a series

• chwala

#### chwala

Gold Member
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...

I mean, do we have to make things complicated here or what?

malawi_glenn
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).
$$2r=f(r+1)-f(r)$$
$$2\sum_{r=1}^nr=\sum_{r=1}^nf(r+1)-\sum_{r=1}^nf(r)=\sum_{r=1}^nf(r+1)-\sum_{r=0}^{n-1}f(r+1)=f(n+1)-f(1)=n(n+1)$$

chwala