Solve the problem involving sum of a series

[chwala]

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...

 

[Hall]

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

 

[anuttarasammyak]

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).

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