<s>[INST] Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
 [/INST] First, show that this is true for $n=1$:
$\left(\sum\limits_{k=1}^{1}k\right)^2=\sum\limits_{k=1}^{1}k^3$
Second, assume that this is true for $n$:
$\left(\sum\limits_{k=1}^{n}k\right)^2=\sum\limits_{k=1}^{n}k^3$
Third, prove that this is true for $n+1$:
$\left(\sum\limits_{k=1}^{n+1}k\right)^2=$
$\left(\color\green{\left(\sum\limits_{k=1}^{n}k\right)}+\color\purple{(n+1)}\right)^2=$
$\color\green{\left(\sum\limits_{k=1}^{n}k\right)}^2+2\color\green{\left(\sum\limits_{k=1}^{n}k\right)}\color\purple{(n+1)}+\color\purple{(n+1)}^2=$
$\color\red{\left(\sum\limits_{k=1}^{n}k\right)^2}+2\left(\sum\limits_{k=1}^{n}k\right)(n+1)+(n+1)^2=$
$\left(\color\red{\sum\limits_{k=1}^{n}k^3}\right)+2\left(\sum\limits_{k=1}^{n}k\right)(n+1)+(n+1)^2=$
$\left(\sum\limits_{k=1}^{n}k^3\right)+n(n+1)(n+1)+(n+1)^2=$
$\left(\sum\limits_{k=1}^{n}k^3\right)+(n+1)^3=$
$\sum\limits_{k=1}^{n+1}k^3$

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