# Improve

• Jan 24th 2008, 10:40 PM
SengNee
Improve
Question:
If $f(r)=r(r+1)(r+2)$, find $f(r+1)-f(r)$ and deduce that $\sum_{r=1}^nr^2=\frac{1}{6}n(n+1)(2n+1)$

My solution:
$f(r+1)-f(r)$
$=(r+1)(r+2)(r+3)-r(r+1)(r+2)$
$=(r+1)(r+2)[(r+3)-r]$
$=3(r+1)(r+2)$
$=3r^2+9r+6$

Let,
$u_r= f(r+1)-f(r)$
$u_r=3r^2+9r+6$
$r^2=\frac{1}{3}u_r-3r-2$

$\sum_{r=1}^nr^2$
$=\sum_{r=1}^n(\frac{1}{3}u_r-3r-2)$
$=\frac{1}{3}\sum_{r=1}^nu_r-3\sum_{r=1}^n r-\sum_{r=1}^n 2$
$=\frac{1}{3}[(n+1)( n+2)( n+3)-2(3)]-3[\frac{n}{2}(n+1)]-2n$
$=\frac{(n^2+3n+2)(n+3)}{3}-2-\frac{3n(n+1)}{2}-2n$
$=\frac{(n^3+3n^2+3n^2+9n+2n+6)}{3}-2-\frac{3n^2+3n}{2}-2n$
$=\frac{2n^3+12n^2+22n+12-12-9n^2-9n-12n}{6}$
$=\frac{1}{6}(2n^3+3n^2+n)$
$=\frac{1}{6}n(2n^2+3n+1)$
$=\frac{1}{6}n(n+1)(2n+1) \quad$ Proven

$\color{blue}\text{Anyone have better or shorter solutions, please show me to improve my long steps.}$

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• Jan 25th 2008, 12:46 AM
angel.white
I don't have a better way, but I was following along for the purposes of learning, and I am confused how you got that
$\sum_{r=1}^nu_r=(n+1)( n+2)( n+3)-2(3)$

If you wouldn't mind explaining that step, I would be appreciative.
• Jan 25th 2008, 01:03 AM
SengNee
Quote:

Originally Posted by angel.white
I don't have a better way, but I was following along for the purposes of learning, and I am confused how you got that
$\sum_{r=1}^nu_r=(n+1)( n+2)( n+3)-2(3)$

If you wouldn't mind explaining that step, I would be appreciative.

If, $u_r=f(r+1)-f(r)$

Then, $\sum_{r=1}^nu_r=f(n+1)-f(1)$
• Jan 25th 2008, 01:32 AM
angel.white
Quote:

Originally Posted by SengNee
If, $u_r=f(r+1)-f(r)$

Then, $\sum_{r=1}^nu_r=f(n+1)-f(1)$

I see, I never learned that (or maybe I just forgot it)