# Proof by Induction

• Apr 10th 2011, 11:26 PM
Oiler
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 $n \in \mathbb{Z}$
• Apr 10th 2011, 11:35 PM
veileen
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.
• Apr 23rd 2011, 09:32 PM
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..
• Apr 24th 2011, 03:24 AM
Quote:

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
• Apr 24th 2011, 07:40 AM
Prove It
• Apr 24th 2011, 02:58 PM
Unfortunately Prove It, as cool as it is, (Cool)
you get zero marks for answering the exam question
"Prove using induction that......."
• Apr 24th 2011, 04:23 PM
Oiler
Quote:

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..
• Apr 24th 2011, 04:27 PM