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?

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 $n=1$.
