1. ## Proof by Induction

Hey all,
can anyone show me how I can prove a statement like this by Induction. Thanks.

$1^3+2^3+3^3+...+n^3=\frac{1}{4}n^2(n+1)^2$ for all $n \in \mathbb{Z}$

2. Well, your n is natural unzero. (the sum starts with 1)
Read this: Mathematical induction - Wikipedia, the free encyclopedia

First step - verify if for n=1 that equality is true.
Second step (Induction) - we (you) suppose that k verify the equality and prove that then k+1 verify too.

3. Hey,

Stuck on the Induction step where I must rewrite the left-hand side..
..(1^3+2^3+3^3+...+n^3)+(n^3+1)=1/4*(n^3+1)(n^3+2)^2

EDIT: Should be (1^3+2^3+3^3+...+n^3)+(n+1)=1/4*(n+1)^2((n+1)+1)^2 and is straight forward to prove from there. Thanks for the help..

4. Originally Posted by Oiler
Hey,

Stuck on the Induction step where I must rewrite the left-hand side..
..(1^3+2^3+3^3+...+n^3)+(n^3+1)=1/4*(n^3+1)(n^3+2)^2

EDIT: Should be (1^3+2^3+3^3+...+n^3)+(n+1)=1/4*(n+1)^2((n+1)+1)^2 and is straight forward to prove from there. Thanks for the help..
No,

you need

P(k)

1^3+2^3+3^3+.....+k^3=(1/4)k^2(k+1)^2

P(k+1)

1^3+2^3+3^3+....+k^3+(k+1)^3=(1/4)(k+1)^2(k+2)^2

and so you will be trying to prove that

(1/4)k^2(k+1)^2+(k+1)^3=(1/4)(k+1)^2(k+2)^2

where both sides have a common factor of (k+1)^2

5. Induction is not necessary...

6. Unfortunately Prove It, as cool as it is,
you get zero marks for answering the exam question
"Prove using induction that......."

7. Originally Posted by Archie Meade
No,

you need

P(k)

1^3+2^3+3^3+.....+k^3=(1/4)k^2(k+1)^2

P(k+1)

1^3+2^3+3^3+....+k^3+(k+1)^3=(1/4)(k+1)^2(k+2)^2

and so you will be trying to prove that

(1/4)k^2(k+1)^2+(k+1)^3=(1/4)(k+1)^2(k+2)^2

where both sides have a common factor of (k+1)^2
Since there's a cube on the LHS reducing it to match the RHS suddenly doesn't seem too trivial..

I get stuck showing the LHS = RHS

..(n^k(k+1)^2+4(k+1)^3))/4..

Thanks for the help..

8. Write the LHS as

(1/4)k^2(k+1)^2+(k+1)(k+1)^2

Then divide both sides by (k+1)^2 to get

(1/4)k^2+k+1 = (1/4)(k+2)^2 =(1/4)(k^2+4k+4)

QED.