Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.
 A: Notice $$
\begin{align}
k^3 
&= \color{red}{(k^3 - k)} + \color{blue}{k}\\
&= \color{red}{(k-1)k(k+1)} + \color{blue}{k}\\
&= \frac{\color{red}{(k-1)k(k+1)}(k+2)-(k-2)\color{red}{(k-1)k(k+1)}}{4}
+ \frac{\color{blue}{k}(k+1) - (k-1)\color{blue}{k}}{2}
\end{align}$$
is telescoping. We have
$$\begin{align}
\sum_{k=0}^n k^3 
&= \frac{(n-1)n(n+1)(n+2)}{4} + \frac{n(n+1)}{2}\\
&= \frac{n(n+1)}{2}\left(\frac{(n-1)(n+2)}{2} + 1\right)\\
&= \left(\frac{n(n+1)}{2}\right)^2
\end{align}
$$
A: Consider the polynomial
$$\sum_{k=0}^n(z+1)^k=\frac{(z+1)^{n+1}-1}{z}=(n+1)+\frac{(n+1)n}2z+\frac{(n+1)n(n-1)}6z^2+\frac{(n+1)n(n-1)(n-2)}{24}z^3...$$
when $z=0$, it takes the value $$\color{blue}{\sum_{k=0}^n1=n+1}.$$
Derive on $z$:
$$\sum_{k=0}^nk(z+1)^{k-1}=\frac{(n+1)n}2+\frac{2(n+1)n(n-1)}6z+\frac{3(n+1)n(n-1)(n-2)}{24}z^2...$$
when $z=0$, it takes the value $$\color{blue}{\sum_{k=0}^nk=\frac{(n+1)n}2}.$$
Multiply by $z+1$:
$$\sum_{k=0}^nk(z+1)^{k}=\frac{(n+1)n}2+\frac{3n(n+1)+2(n+1)n(n-1)}6z+\frac{8(n+1)n(n-1)+3(n+1)n(n-1)(n-2)}{24}z^2...$$
Derive on $z$:
$$\sum_{k=0}^nk^2(z+1)^{k-1}=\frac{(n+1)n(2n+1)}6+\frac{(n+1)n(n-1)(3n+2)}{12}z...$$
when $z=0$, it takes the value $$\color{blue}{\sum_{k=0}^nk^2=\frac{(n+1)n(2n+1)}6}.$$
Multiply once more by $z+1$ and derive to get the third order sum. The constant term is
$$\frac{(n+1)n(2n+1)}6+\frac{(n+1)n(n-1)(3n+2)}{12}=\frac{3(n+1)n(n^2+n)}{12},$$
giving:
$$\color{blue}{\sum_{k=0}^nk^3=\frac{(n+1)^2n^2}4}.$$
A: Hint: If you know the closed forms of $\displaystyle\sum_{k=0}^n1$, $\displaystyle\sum_{k=0}^nk$, and $\displaystyle\sum_{k=0}^nk^2$, use the fact that
\begin{align*}
(n+1)^4 &= \sum_{k=0}^n(k+1)^4 - \sum_{k=0}^nk^4\\ 
&= \sum_{k=0}^n(k^4+4k^3+6k^2+4k+1) - \sum_{k=0}^nk^4\\
&= 4\sum_{k=0}^nk^3 + 6\sum_{k=0}^nk^2+4\sum_{k=0}^nk + \sum_{k=0}^n1.
\end{align*}
A: $\sum\limits_{k=0}^{n}k^3=\frac{n^2(n+1)^2}{4}$.
Prove it by induction!
Alternatively, to find this, we can use the identity $$(1^3+2^3+3^3+\cdots+n^3)\equiv(\color{green}{1+2+3+\cdots+n})^2.$$
Now, $\color{green}{1+2+3+\cdots+n} \equiv \sum\limits_{k=0}^{n}k\equiv\frac{n(n+1)}{2}$
A: First, we know that: $$\sum_{i \mathop = 1}^n i = \frac {n \left({n + 1}\right)} 2.$$ Thus: $$\left({\sum_{i \mathop = 1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4.$$ Next we use induction on $n$ to show that $$\sum_{i \mathop = 1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4.$$
The base case holds since $$1^3 = \frac{1 \left({1 + 1}\right)^2} 4.$$
Now we need to show that if it holds for $n$, then it holds for $n+1$. $$\eqalign{
\sum_{i \mathop = 1}^n i^3 + \left({n + 1}\right)^3 &= \sum_{i \mathop = 1}^n i^3 + \left({n + 1}\right)^3 \\
&= \frac{n^2 \left({n + 1}\right)^2} 4 + \left({n + 1}\right)^3 \\
&= \frac{n^4 + 2 n^3 + n^2} 4 + \frac {4 n^3 + 12 n^2 + 12 n + 4} 4 \\
&=  \frac{n^4 + 6 n^3 + 13 n^2 + 12 n + 4} 4 \\
&=  \frac{\left({n + 1}\right)^2 \left({n + 2}\right)^2} 4.
}$$
So by induction we have that: $$\boxed{\ \displaystyle\sum_{i \mathop = 1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4. \ }$$
