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?

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 $f(0)=0$ 
