If f(r) = r(r+1)(r+2), simplify f(r+1)-f(r). Hence find ∑_(r=1)^n▒〖3(r^2+3r+2)〗and deduce ∑_(r=1)^n▒r^2. I stuck at the deduction part, can anyone explain how to do it?
Since f(r) = r(r+1)(r+2), have f(r+1)-f(r) = (r+1)((r+1)+1)((r+1)+2) - r(r+1)(r+2) = (r+1)(r+2)(r+3) - r(r+1)(r+2) = (r+1)(r+2){(r+3)-(r)} = 3(r+1)(r+2).
Thus $\displaystyle f(r+1)-f(r) = 3(r^{2}+3r+2)$.
$\displaystyle \sum_{r=1}^{n} 3(r^{2}+3r+2) = \sum_{r=1}^{n} ( f(r+1) - f(r) ) = f(n+1) - f(1) $ (the last is because the sum telescopes - everything else cancels out.)
Now get f(n+1) - f(1): $\displaystyle f(n+1) - f(1) = (n+1)(n+2)(n+3) - (6) = n^{3} + 6n^{2} + 11n + 6 - 6$
Thus $\displaystyle f(n+1) - f(1) = n(n^{2} + 6n + 11) $
You're presumed to already know that $\displaystyle \sum_{r=1}^{n} 1 = n$, and $\displaystyle \sum_{r=1}^{n} r = n(n+1)/2$.
Now $\displaystyle \sum_{r=1}^{n} 3(r^2+3r+2) = 3 ( \sum_{r=1}^{n} r^{2} ) + 9 ( \sum_{r=1}^{n} r ) + 6 ( \sum_{r=1}^{n} 1 )$
So pluggin in gives: $\displaystyle ( f(n+1) - f(1) ) = 3 ( \sum_{r=1}^{n} r^{2} ) + 9 ( n(n+1)/2 ) + 6 ( n )$
Solving for our desired series, $\displaystyle \sum_{r=1}^{n} r^{2}$, gives:
$\displaystyle \sum_{r=1}^{n} r^{2} = (1/3)( f(n+1) - f(1) - 9n(n+1)/2 - 6n )$
$\displaystyle = (1/3)( n(n^{2} + 6n + 11) - 9n(n+1)/2 - 6n )$
$\displaystyle = (n/3)( n^{2} + 6n + 11 - 9(n+1)/2 - 6 )$
$\displaystyle = (n/3)( n^{2} + (3/2)n + 1/2 )$
$\displaystyle = (n/6)( 2n^{2} + 3n + 1 )$
$\displaystyle = (n/6)( ( n + 1 ) ( 2n + 1 ) )$
$\displaystyle = ( n )( n + 1 )( 2n + 1 )/6$
Thus $\displaystyle \sum_{r=1}^{n} r^{2} = ( n )( n + 1 )( 2n + 1 ) / 6$