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?
 A: Induction on k.
Base Case: 1=1
Induction Hypothesis: $(1+2+{...}+k)^2$=$1^3 + {...} + k^3$
Based on the hypothesis: 
$(1+2+{...}+k+(k+1))^2$
=$(1+2+{...}+k)^2$+2$(1+2+{...}+k)(k+1)$+$(k+1)^2$
=$1^3 + {...} + k^3$+$k(k+1)^2$+$(k+1)^2$
=$1^3 + {...} + k^3 + (k+1)^3$
Done.
A: 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.
A: Let $T_n=\sum_{k=0}^nk=\frac{n(n+1)}2$ be the $n^{th}$ triangular number. Then,
$$T_n^2-T_{n-1}^2=(T_n-T_{n-1})(T_n+T_{n-1})=n(n+T_{n-1}+T_{n-1})=n^3.$$

Base:
$$T_1^2=(0+1)^2=1=(0)^2+1^3=T_0^2+1^3.$$
Induction:
$$T_n^2=T_{n-1}^2+n^3.$$
A: We have:
$$(1+2+...+k+(k+1))^2=(1+2+...+k)^2+(k+1)^2+2(k+1)(1+2+...+k)$$
So by induction hypothesis:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+2(k+1)(1+2+...+k)$$
We know $1+2+...+k = \frac{k(k+1)}{2}$.(You can prove it by induction easily). Therefore:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+k(k+1)^2=1^3+2^3+...+k^3+(k+1)^3$$
A: We have $(a+b)^2=a^2+b^2 + 2ab.$ Thus $$(1+2+...+(k+1))^2=(1+2+...+k)^2+(k+1)^2+ 2(1+2+...+k)(k+1)=1^3+...+k^3+(k+1)^2 + 2(1+2+...+k)(k+1)$$ 
Using $1+...+k=\dfrac{k(k+1)}{2}$ we get $$(1+2+...+k+1)^2=1^3+...+k^3+ (k+1)^2 + k(k+1)^2=1^3+...+k^3+ (k+1)^2(k+1)=1^3+...+(k+1)^3$$ and the inductive step is proven.
A: After checking this for k=1, k=2 you need to follow this:
Let's say $A_1=(1+2+...+k)^2$; $B_1=1^3+...k^3$ and $A_2=(1+2+...+ (k+1))^2$
$B_2=1^3+...(k+1)^3$
Then to use induction it is needed to prove that $A_2-A_1=B_2-B_1$ or:
$$(1+2+...+ (k+1))^2-(1+2+...+k)^2=1^3+...(k+1)^3-1^3+...k^3$$
And it is equal to:
$$(k+1)(1+2+...+k)+(1+2+...+k+1)=(k+1)^3$$
$$(k+1)(\frac{k(k+1)}{2}+\frac{(k+1)(k+2)}{2})=(k+1)^3$$
$$(k+1)^2\frac{2k+2}{2}=(k+1)^3$$
$$(k+1)^3=(k+1)^3$$
This is the prove.
