Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$
To be proven: $((k)(k+1)/2)^2 + (k+1)^3 = ((k+1)(k+2)/2)^2$
My problem is that I do not know how I can put the $ + (k+1)^3$ inside $((k)(k+1)/2)^2$. 
Simplifying the left and right part of the statement does not help:
Simplifying the left side: $((k)(k+1)/2)^2 + (k+1)^3 = ((k^2+k)/2)^2 + (k+1)^3 $ 
Simplifying the right side: $((k+1)(k+2)/2)^2 = ((k^2+3k+2)/2)^2$
So i am left with: $((k^2+k)/2)^2 + (k+1)^3 = ((k^2+3k+2)/2)^2$
That is the same as: $1/4 (k^2+k)^2 + (k+1)^3 = 1/4((k^2+3k+2))^2$
Going further with the left side:$1/4 (k^2+k)^2 + (k+1)^3 = (1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1$
Going further with the right side: $1/4((k^2+3k+2))^2 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am stuck with: $(1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am kind of left with garbage. Am I missing something? What do I do wrong? Where can I find a good resources to learn how to solve this issue?
 A: Show that $S_n-S_{n-1}=n^3$ (or if you prefer, $S_n=S_{n-1}+n^3$).
Indeed,
$$\left(\frac{n(n+1)}2\right)^2-\left(\frac{(n-1)n}2\right)^2=n^2\left(\frac{n+1}2+\frac{n-1}2\right)\left(\frac{n+1}2-\frac{n-1}2\right).$$

Note that induction is unnecessary. As both members are quartic polynomials, it suffices to prove equality for five distinct values of $n$:
$$0=0,\\1=1,\\9=3^2,\\36=6^2,\\100=10^2.$$
A: Hint
It's easier to begin by the end 
$$\left(\frac{(n+1)(n+2)}{2}\right)^2-\frac{n(n+1)}{2}^2=$$
$$\left(\frac{n+1}{2}\right)^2\left((n+2)^2-n^2\right)=$$
$$\left(\frac{n+1}{2}\right)^2(4(n+1))=$$
$$(n+1)^3.$$
qed.
A: Take your "to be proven" equation and divide through by $(k+1)^2$ and multiply through by $4$. This gives $k^2+4(k+1)=(k+2)^2$ as the equation to be proven, and this is trivial.
A: \begin{align}
\underbrace{1^3+2^3+\ldots+n^3}_{\left[\frac{n(n+1)}{2} \right]^2}+(n+1)^3
=&
\left[\frac{n(n+1)}{2} \right]^2+(n+1)^3
\\
=&
\frac{n^2\color{red}{(n+1)^2}}{4}
+
(n+1)\color{red}{(n+1)^2}
\\
=&
\left\lgroup 
\frac{n^2}{4}
+
(n+1)
\right\rgroup
\color{red}{(n+1)^2}
\\
=& \left\lgroup \frac{n^2 +4(n+1)}{4} \right\rgroup \color{red}{(n+1)^2}\\ 
=&\left\lgroup
\frac{(n+2)^2}{4}
\right\rgroup \color{red}{(n+1)^2}
\\
=&\left\lgroup
\frac{(n+2)}{2}
\right\rgroup^2 \color{red}{(n+1)^2}
\\
=&\left\lgroup
\frac{\big(\color{red}{(n+1)}+1\big)}{2}
\right\rgroup^2 \color{red}{(n+1)^2}
\\
=&\left[
\frac{\color{red}{(n+1)}\big(\color{red}{(n+1)}+1\big)}{2}
\right]^2
\end{align}
A: After your $" to be proven "$ step, follow the rules below: 
1. Take $ (k+1)^{2} $ common so we have then $ (k+1)+\frac {k}{4} ^{2} $.
2. Then simplify the result to get $ \frac { k^{2}+4k+4}{4} $.
3.This is equivalent to $\frac {(k+2)}{2}^{2} $
4. Proof finished as we have our result equal to $ \frac {(k+1)(k+2)}{2} ^{2} $. 
A: First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^3=(1(1+1)/2)^2$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^3=(n(n+1)/2)^2$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^3=$
$\color\red{\sum\limits_{k=1}^{n}k^3}+(n+1)^3=$
$\color\red{(n(n+1)/2)^2}+(n+1)^3=$
$((n+1)(n+2)/2)^2$

Please note that the assumption is used only in the part marked red.
