Hey all,

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

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

Printable View

- Apr 10th 2011, 11:26 PMOilerProof by Induction
Hey all,

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

$\displaystyle 1^3+2^3+3^3+...+n^3=\frac{1}{4}n^2(n+1)^2$ for all $\displaystyle n \in \mathbb{Z}$ - Apr 10th 2011, 11:35 PMveileen
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 PMOiler
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 AMArchie Meade
- Apr 24th 2011, 07:40 AMProve It
Induction is not necessary...

http://i22.photobucket.com/albums/b3...sumofcubes.jpg - Apr 24th 2011, 02:58 PMArchie Meade
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 PMOiler
- Apr 24th 2011, 04:27 PMArchie Meade
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.