Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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: $$T_n^2=T_{n-1}^2+n^3.$$

Answer 4

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: $$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+k(k+1)^2=1^3+2^3+...+k^3+(k+1)^3$$

Answer 5

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.

Answer 6

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.