Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$

Question

Question:

Given that $\displaystyle\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$

Attempt:

Substituting $n = -2,-1,0,1,2$into $\sum_{r=-1}^{n}{r^3}$ we get: $$\sum_{r=-2}^{-2}{r^3} = -8 =16a-8b+4c-2d+e$$ $$\sum_{r=-2}^{-1}{r^3} = -9 = a-b+c-d+e$$ $$\sum_{r=-2}^{0}{r^3} = -9 = e$$ $$\sum_{r=-2}^{1}{r^3} = -8 = a+b+c+d+e$$ $$\sum_{r=-2}^{2}{r^3} = 0 =16a+8b+4c+2d+e$$ After some algebra we now know $a= -\frac12, b = -1, c = 1, d = \frac32, e = -9$: $$\therefore \sum^{n}_{r=-2}{r^3}= -\frac12n^4 -n^3 + n^2 +\frac32n -9$$ We define $\sum^{n}_{r=0}{r^3}$: $$\sum^{n}_{r=0}{r^3}=\sum^{n}_{r=-2}{r^3} - \sum^{-2}_{r=-2}{r^3}=-\frac12n^4 -n^3 + n^2 +\frac32n -9 -(-9)$$ $$-\frac12n^4 -n^3 + n^2 +\frac32n = \frac14(-2n^4-4n^3+4n^2 + 6n)$$ $$\frac14n^2\biggl(-2n^2-4n+4+\frac6n\biggl) = \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$ $$ \frac14n^2\biggl(\frac{-2n^3-4n^2+4n+6}{n}\biggl)$$

My problem:

I got $d = \frac32$ and as far as i can see since it need to end up with $d = 0$ as the smallest term that can be left is a $n^2$ term as $\frac14n^2(n+1)^2$ is multiplied by $n^2$ meaning $e$ has to cancel out (which it does) but i don't know how to get rid of $d$, if you could explain it to me it would be much appreciated

Answer 1

I would propose a more efficient strategy. Given that $\sum_{r=-2}^{n}r^3$ is a fourth-degree polynomial in the variable $n$, the same holds for $\sum_{r=0}^{n}r^3=\sum_{r=1}^{n}r^3$, since the two sums differ by a constant. Two different fourth-degree polynomials cannot agree on $5$ points or more: otherwise their difference would be a non-constant polynomial with degree $\leq 4$ and with $\geq 5$ roots, impossible.
So, in order to prove

$$ \sum_{r=0}^{n}r^3 = \frac{n^2(n+1)^2}{4} \tag{1}$$ for any $n\in\mathbb{N}$, it is enough to check that such identity holds at $n\in\{0,1,2,3,4\}$:

$$ \begin{array}{|c|c|c|c|c|c|}\hline n & 0 & 1 & 2 & 3 & 4 & 5\\ \hline \sum_{r=0}^{n}r^3 & 0 & 1 & 9 & 36 & 100 & 225\\ \hline \frac{1}{4}\left(n(n+1)\right)^2 & 0 & 1 & 9 & 36 & 100 & 225 \\\hline \end{array}\tag{2}$$ and we are done.

Answer 2

for $n \ge k$ let the sum $\sum_{r=k}^n r^3 $ be $S_{k}(n)$ so for $n \ge 0$ $$ S_{-2}(n) = S_0(n) -9 $$ since $S_0(0)=0$ this gives $S_{-2}(0)=e=-9$ and $$ S_0(n)= an^4+bn^3+cn^2+dn $$ now $S_0(n)-S_0(n-1) = n^3$ so $$ \begin{align} a(4n^3-6n^2+4n-&1) + \\ b(3n^2-3n+&1)+\\ c(2n-&1)+\\ &d= n^3 \end{align} $$ comparing coefficients of $n^3$ gives $a=\frac14$. the $n^2$ term gives $b=2a$ so $b=\frac12$. the term in $n$ gives $4a-3b+2c=0$ so $c-\frac14$. and for the constant term: $a-b+c-d=0$ so $d=0$. altogether this gives $$ S_0(n)=\frac14 n^4 + \frac12 n^3 + \frac14 n^2 = \bigg(\frac{n(n+1)}2 \bigg)^2 $$

Answer 3

$$\begin{align} &n=0: &e&=-9\tag{1}\\\\ &n=1: &a+b+c+d&=1\tag{2}\\ &n=-1: &a-b+c-d&=0\tag{3}\\\\ \frac 12[(2)+(3)]: &&a+c&=\frac 12\tag{4}\\ \frac 12[(2)-(3)]: &&b+d&=\frac 12\tag{5}\\\\ &n=2: &16a+8b+4c+2d&=9\tag{6}\\ &n=-2: &16a-8b+4c-2d&=1\tag{7}\\\\ \frac 12 [(6)+(7)]: &&16a+4c&=5 \quad\Rightarrow a=\frac 14, c=\frac 14 \;\;\ \text{using }(4)\\ \frac 14[(6)-(7)]: &&4b+d&=2 \quad\Rightarrow b=\frac 12, d=0 \quad\text{using }(5)\\\\ &&\sum_{r=0}^n r^3&=\left(\sum_{r=-2}^n r^3\right) -9\\ && &=an^4+bn^3+cn^2+dn+e-(-9)\\ && &=\frac 14 n^2+\frac 12 n^3+\frac 14 n^2\\ && &=\frac 14n^2(n+1)^2\\ && &=\left(\frac 12 n(n+1)\right)^2\end{align}$$

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See also this question I posted.

If we can show that $n^2$ and $(n+1)^2$ are factors of the sum of odd powers $p(>1)$ of the first $n$ integers, then for $p=3$ (i.e. sum of cubes) we would just have to determine the constant ($\frac 14$, by putting $n=1$) and we are done.

Answer 4

I finished it so here's the full version if anyone is interested:

It is given that $\sum^{n}_{r=-1}{r^2}$ can be written in the form $pn^3 + qn^2 + rn +s$, where $p, q, r$ and $s$ are numbers. By setting $n = -1, 0, 1$ and $2$, obtain four equations that must be satisfied by $p, q, r$ and $s$ and hence show that:$$\sum_{r=0}^{n}{r^2} = \frac16n(2n+1)(n+1)$$ Given that $\sum^{n}_{r=-2}{r^3}$ can be written in the form $an^4+bn^3+cn^2+dn+e$, show similarly that: $$\sum^{n}_{r=0}{r^3}=\frac14n^2(n+1)^2$$

Substituting $n = -1,0,1,2$ into $\sum_{r=-1}^{n}{r^2}$ we get: $$\sum_{r=-1}^{-1}{r^2} = 1 =-p+q-r+s$$ $$\sum_{r=-1}^{0}{r^2} = 1 = s$$ $$\sum_{r=-1}^{1}{r^2} = 2 = p+q+r+s$$ $$\sum_{r=-1}^{2}{r^2} = 6 = 8p+4q+2r+s$$ Now sub $s=1$ into our 3 equations: $$1 = -p+q-r+1\Rightarrow 0 = -p+q-r$$ $$ 2 = p +q + r + 1 \Rightarrow 1 = p+q+r$$ $$ 6 = 8p+4q+2r+1 \Rightarrow 5 = 8p + 4q+ 2r$$ Now we obtain $q$: $$0+1 = (-p+q-r)+(p+q+r)$$ $$1= (1-1)p+(1+1)q+(1-1)r \Rightarrow q = \frac{1}{2}$$ Now sub $q = \frac12$ into our 2 equations: $$0 = -p+\frac12-r\Rightarrow \frac12 = p+r$$ $$ 5 = 8p+\frac42+2r \Rightarrow 3 = 8p + 2r$$ Now find $p$: $$\frac12 = p + r \Rightarrow 1 = 2p + 2r$$ $$3 - 1 = (8p + 2r) - (2p + 2r)$$ $$2 = (8-2)p - (2-2)r\Rightarrow p = \frac26$$ Now sub $p$ to find $r$: $$\frac12 = \frac26 + r \Rightarrow r = \frac16$$ We now know $p = \frac26,q = \frac12, r=\frac16, s = 1$: $$\therefore \sum_{r=-1}^{n}{r^2} = \frac26n^3 + \frac12n^2 + \frac16n + 1$$ We define $\sum_{r=0}^{n}{r^2}$ $$\sum_{r=0}^{n}{r^2} = \sum_{r=-1}^{n}{r^2} - \sum_{r=-1}^{-1}{r^2} = \biggl( \frac26n^3 + \frac12n^2 + \frac16n + 1\biggr) - (-1)^2$$ $$\frac26n^3 + \frac12n^2 + \frac16n + 1 - 1 = \frac26n^3 + \frac12n^2 + \frac16n$$ Now simplify $\sum_{r=0}^{n}{r^2}$: $$\frac26n^3 + \frac12n^2 + \frac16n = \frac16(2n^3+3n^2+n)$$ $$\frac16n(2n^2+3n+1) = \frac16n(2n+1)(n+1) $$ $$\therefore \sum_{r=0}^{n}{r^2} = \frac16n(2n+1)(n+1)$$ \ Substituting $n = -2,-1,0,1,2$into $\sum_{r=-1}^{n}{r^3}$ we get: $$\sum_{r=-2}^{-2}{r^3} = -8 =16a-8b+4c-2d+e$$ $$\sum_{r=-2}^{-1}{r^3} = -9 = a-b+c-d+e$$ $$\sum_{r=-2}^{0}{r^3} = -9 = e$$ $$\sum_{r=-2}^{1}{r^3} = -8 = a+b+c+d+e$$ $$\sum_{r=-2}^{2}{r^3} = 0 =16a+8b+4c+2d+e$$ Now sub $e = -9$ into our $3$ equations: $$-8 = 16a-8b+4c-2d-9\Rightarrow 1 = 16a-8b+4c-2d$$ $$-9 = a-b+c-d-9\Rightarrow 0 = a-b+c-d$$ $$-8 = a+b+c+d-9\Rightarrow 1 = a+b+c+d$$ $$0 =16a+8b+4c+2d - 9 \Rightarrow 9 = 16a+8b+4c+2d$$ Now we obtain $c$: $$0+1 = (a-b+c-d)+(a+b+c+d)$$ $$1 = (1+1)a+(1-1)b+(1+1)c+(1-1)d \Rightarrow 2a = 1- 2c$$ $$1+9 = (16a-8b+4c-2d)+(16a+8b+4c+2d)$$ $$10 = (16+16)a+(8-8)b+(4+4)c+(2-2)d \Rightarrow 32a + 8c = 10$$ $$16(2a) + 8c = 16(1-2c) +8c = -24c +16 = 10 \Rightarrow c = \frac14$$ Now sub in $c = \frac14$ to obtain $a$: $$2a = 1 - 2(\frac14) \Rightarrow a = \frac{1}{2}$$ Now sub $c =\frac14, a= \frac14$ into our $2$ equations: $$0 = \frac14-b+\frac14-d \Rightarrow \frac12 = b + d$$ $$1 = 16(\frac14)-8b+4(\frac14)-2d \Rightarrow 4 = 8b + 2d$$ Now we obtain $b$: $$4-2(\frac12) = 8b + 2d -2(b+d)$$ $$3 = (8-2)b + (2-2)d \Rightarrow b = \frac12$$ Now sub in $b = \frac12$ to obtain $d$: $$4 = 8(\frac12) + 2d \Rightarrow 0 = d$$ We now know $a= \frac14, b = \frac12, c = \frac14, d = 0, e = -9$: $$\therefore \sum^{n}_{r=-2}{r^3}= \frac14n^4 + \frac12n^3 + \frac14n^2 -9$$ We define $\sum^{n}_{r=0}{r^3}$: $$\sum^{n}_{r=0}{r^3}=\sum^{n}_{r=-2}{r^3} - \sum^{-2}_{r=-2}{r^3}=\frac14n^4 +\frac12n^3 + \frac14n^2 -9 -(-9)$$ $$\frac14n^4 +\frac12n^3 + \frac14n^2 = \frac14 (n^4+2n^3+n^2)$$ $$\frac14n^2(n^2+2n+1) = \frac14n^2(n+1)^2$$ $$\therefore \sum^{n}_{r=0}{r^3}= \frac14n^2(n+1)^2$$

(thanks for all the answers and help)