Question:

If $\displaystyle f(r)=r(r+1)(r+2) $, find $\displaystyle f(r+1)-f(r)$ and deduce that $\displaystyle \sum_{r=1}^nr^2=\frac{1}{6}n(n+1)(2n+1)$

My solution:

$\displaystyle f(r+1)-f(r) $

$\displaystyle =(r+1)(r+2)(r+3)-r(r+1)(r+2) $

$\displaystyle =(r+1)(r+2)[(r+3)-r] $

$\displaystyle =3(r+1)(r+2) $

$\displaystyle =3r^2+9r+6$

Let,

$\displaystyle u_r= f(r+1)-f(r) $

$\displaystyle u_r=3r^2+9r+6$

$\displaystyle r^2=\frac{1}{3}u_r-3r-2$

$\displaystyle \sum_{r=1}^nr^2$

$\displaystyle =\sum_{r=1}^n(\frac{1}{3}u_r-3r-2) $

$\displaystyle =\frac{1}{3}\sum_{r=1}^nu_r-3\sum_{r=1}^n r-\sum_{r=1}^n 2$

$\displaystyle =\frac{1}{3}[(n+1)( n+2)( n+3)-2(3)]-3[\frac{n}{2}(n+1)]-2n$

$\displaystyle =\frac{(n^2+3n+2)(n+3)}{3}-2-\frac{3n(n+1)}{2}-2n$

$\displaystyle =\frac{(n^3+3n^2+3n^2+9n+2n+6)}{3}-2-\frac{3n^2+3n}{2}-2n$

$\displaystyle =\frac{2n^3+12n^2+22n+12-12-9n^2-9n-12n}{6}$

$\displaystyle =\frac{1}{6}(2n^3+3n^2+n) $

$\displaystyle =\frac{1}{6}n(2n^2+3n+1) $

$\displaystyle =\frac{1}{6}n(n+1)(2n+1) \quad $ Proven

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

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