Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$ The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$
[Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]
My solution:
$$\begin{align}
\sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\
&=\frac{n}{2}[\text{first term} + \text{last term}]\\
&=\frac{n(1^3+n^3)}{2}
\end{align}$$
What am I doing wrong?
 A: Let's take the suggested hint, and consider $$(k+1)^{4}-k^{4}=4k^{3}+6k^{2}+4k+1$$
Summing up both sides from $k=1$ to $n$. Notice that $$\sum_{k=1}^{n}[(k+1)^{4}-k^{4}]=[2^{4}-1^{4}]+[3^{4}-2^{4}]+\ldots+[n^{4}-(n-1)^{4}]+[(n+1)^{4}-n^{4}]$$
Cancelling, we get $(n+1)^{4}-1$. So altogether,
$$(n+1)^{4}-1=4\sum_{k=1}^{n}k^{3}+6\sum_{k=1}^{n}k^{2}+4\sum_{k=1}^{n}k+\sum_{k=1}^{n}1$$
Assuming you already know the results for $\sum1$, $\sum k$, $\sum k^{2}$, we can substitute them in:
$$n^{4}+4n^{3}+6n^{2}+4n+1-1=4S+n(n+1)(2n+1)+2n(n+1)+n$$
$$[n^{4}+4n^{3}+6n^{2}+4n]-[2n^{3}+3n^{2}+n]-[2n^{2}+2n]-[n]=4S$$
Thus, simplifying further
$$n^{4}+2n^{3}+n^{2}=n^{2}(n+1)^{2}=4S \implies \sum_{k=1}^{n}k^{3}=\frac{n^{2}(n+1)^{2}}{4}  \square$$
A: Method 1
Using the binomial identity
$$
\sum_{k=m}^{n-j}\binom{n-k}{j}\binom{k}{m}=\binom{n+1}{j+m+1}\tag{1}
$$
with $j=0$ yields
$$
\sum_{k=0}^n\binom{k}{m}=\binom{n+1}{m+1}\tag{2}
$$
Using $(2)$ and the identity
$$
k^3=6\binom{k}{3}+6\binom{k}{2}+\binom{k}{1}\tag{3}
$$
we get that
$$
\begin{align}
\sum_{k=0}^nk^3
&=6\binom{n+1}{4}+6\binom{n+1}{3}+\binom{n+1}{2}\\
&=6\binom{n}{4}+12\binom{n}{3}+7\binom{n}{2}+\binom{n}{1}\\
&=\frac{n^2(n+1)^2}{4}\tag{4}
\end{align}
$$

Method 2
Using the Euler-Maclaurin Sum Formula, we get
$$
\sum_{k=0}^nk^3=\frac14n^4+\frac12n^3+\frac14n^2+C\tag{5}
$$
where we get $C=0$ by plugging in $n=1$.
A: $$(k+1)^4 = k^4 + 4k^3 + 6k^2 + 4k + 1$$
Take sum from k=1 to n
$$\sum_{k=1}^n(k+1)^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$
$$\sum_{k=2}^{n+1}k^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n $$
Add 1 on both sides
$$\sum_{k=1}^{n}k^4 + (n+1)^4 = \sum_{k=1}^nk^4 + \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n + 1$$
Cancel $\sum_{k=1}^{n}k^4$
$$ (n+1)^4 =  \sum_{k=1}^n4k^3 + \sum_{k=1}^n6k^2 + \sum_{k=1}^n4k + n + 1$$
You need $\sum_{k=1}^n k^2, \sum_{k=1}^n k$ to solve it. Hope you can do the rest.
A: For a geometric solution, you can see theorem 3 on the last page of this PDF. Sorry I did not have time to type it here. This solution was published by Abu Bekr Mohammad ibn Alhusain Alkarachi in about A.D. 1010 (Exercise 40 of appendix E, page A38 of Stewart Calculus 5th edition).
A: Let $f(n)=\Big(\frac{n(n+1)}2\Big)^2$ , then$\space$$f(n-1)=\Big(\frac{n(n-1)}2\Big)^2$ ; now
 we know that 
$(n+1)^2 - (n-1)^2=4n$ , 
this implies 
$\space$ $n^2\Big((n+1)^2 - (n-1)^2\Big)=4n^3=$$\big(n(n+1)\big)^2 - \big(n(n-1)\big)^2$ 
$\implies$ $\frac{\big(n(n+1)\big)^2}4 - \frac{\big(n(n-1)\big)^2}4=$ $\Big(\frac{n(n+1)}2\Big)^2-\Big(\frac{n(n-1)}2\Big)^2=n^3$$=f(n)-f(n-1)$ 
$\implies$   $\sum_{n=1}^m \Big(f(n)-f(n-1)\Big) =$$\sum_{n=1}^mn^3$$=f(m)-f(0)=f(m)=\Big(\frac{m(m+1)}2\Big)^2$ , where we 
have used $f(0)=0$ 
