If $f(r)=r(r+1)(r+2)(r+3)$, simplify $f(r+1)-f(r)$ and use your result to find $$\sum_{r=1}^n r^3$$
I need help with the second part of the question because the book does not give any answers for the second part.
I have found that
$f(r+1)-f(r)$
$=(r+1)(r+2)(r+3)(r+4)-r(r+1)(r+2)(r+3)$
$=(r+1)(r+2)(r+3)(r+4-r)$
$=4(r+1)(r+2)(r+3)$
I am certain that $f(r)=r(r+1)(r+2)(r+3) =4(r+1)(r+2)(r+3)$ because that is the answer given by the book.
To find $$\sum_{r=1}^n r^3$$ I solved for $$\sum_{r=1}^n [4(r+1)(r+2)(r+3)]$$
$$=\sum_{r=1}^n [(r+1)(r+2)(r+3)(r+4)-r(r+1)(r+2)(r+3)]$$ $=(n+1)(n+2)(n+3)(n+4)-24$
Therefore,
$$\sum_{r=1}^n [4r^3+24r^2+44r+24]=(n+1)(n+2)(n+3)(n+4)-24$$ $$\sum_{r=1}^n 4r^3+ \sum_{r=1}^n24r^2 + \sum_{r=1}^n44r + \sum_{r=1}^n 24=(n+1)(n+2)(n+3)(n+4)-24$$
$$\begin{align} 4 \sum_{r=1}^nr^3&=(n+1)(n+2)(n+3)(n+4)-24-24[\frac{n}{2}(n+1)]^2 -44[\frac{n}{2}(n+1)]-24n\\ 4 \sum_{r=1}^nr^3&=(n+1)(n+2)(n+3)(n+4)-24-24[\frac{n^2}{4}(n+1)^2]-44[\frac{n}{2}(n+1)]-24n\\ 4 \sum_{r=1}^nr^3&=(n+1)(n+2)(n+3)(n+4)-24-6n^2(n+1)^2-22n(n+1)-24n\\ 4 \sum_{r=1}^nr^3&=(n^3+6n^2+11n+6)(n+4)-24-6n^2(n^2+2n+1)-22n(n+1)-24n\\ 4 \sum_{r=1}^nr^3&=n^4+10n^3+35n^2+50n+24-24-6n^4-12n^3-6n^2-22n^2-22n-24n\\ 4 \sum_{r=1}^nr^3&=-5n^4-2n^3+7n^2+4n \end {align}$$
I know that summation of a finite series should be positive, but the book does not give a solution for $$\sum_{r=1}^n r^3$$
So what went wrong?
As J. W. Tanner's question comment indicated, you made a mistake in the third line of your set of equations, i.e., in the third term on the right side of
$$4 \sum_{r=1}^nr^3=(n+1)(n+2)(n+3)(n+4)-24-24[\frac{n^2}{4}(n+1)^2]-44[\frac{n}{2}(n+1)]-24n \tag{1}\label{eq1A}$$
You are using that
$$\sum_{r=1}^{n}r^2 = \frac{n^2}{4}(n+1)^2 \tag{2}\label{eq2A}$$
This is actually the sum for the cubes, which you are trying to find, as you can see in Faulhaber's formula. Instead, the sum for the squares, as shown in that same Wikipedia article, is
$$\sum_{r=1}^{n}r^2 = \frac{n(n+1)(2n+1)}{6} = \frac{2n^3 + 3n^2 + n}{6} \tag{3}\label{eq3A}$$