Prove that $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$

Question

I know that it is true but not sure how to write the proof for: $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$. I need help to guide me in the right direction.

Thanks in advance.

edit:

Okay at n=k I have $ 1+4+9 ... + k^2 = (k/6)(k+1)(2k+1)$

and at $n=k+1$ I have $((k+1)/6))((k+1)+1)(2(k+1)+1) = ((k+1)/6)(k+2)(2k+3).$

Does my base step need to begin at $n=0$ or $n=1?$ How do I tell?

Well first I'm trying to figure out if I this is something I can prove by induction. For instance is it necessary to have a base step, the assumption, and the n+1 step in this proof? @PaulSundheim — Slae, Oct 15 '14 at 03:16
In the second proof below, assume that the theorem is true for some $n=k$. The rest of the proof works fine though. — Paul Sundheim, Oct 15 '14 at 03:19

score 3 · Answer 1 · answered Oct 15 '14 at 03:17

3

Hint:

$(n+1)^3-n^3=3n^2+3n+1$,

and then sum it from 1 to $n$.

answered Oct 15 '14 at 03:17

Robert Fan

759

score 1 · Answer 2 · answered Dec 15 '14 at 05:15

1

Proof by picture $%random text fdfdsfdsfdsfdsfdsfs$:

enter image description here

answered Dec 15 '14 at 05:15

Bruno Joyal

54,711

score 1 · Accepted Answer · answered Oct 15 '14 at 03:17

1

When $n=1$, it is Okay.

Suppose that $n=k$, $$1 + 4 + 9 \cdots + k^2 = (k/6)(k+1)(2k+1).$$ Then $n=k+1$, $$1 + 4 + 9 \cdots + k^2+(k+1)^2 = (k/6)(k+1)(2k+1)+(k+1)^2=((k+1)/6)(k+2)(2k+3).$$

answered Oct 15 '14 at 03:17

Paul

20,553

score 0 · Answer 4 · edited Dec 15 '14 at 04:52

0

There is an answer from U of Manitoba at the following link: Link to proof

In the future, type your problem in Google Search to see if the solution is out there already. Odds are, the answer to your question is already there. I found it with ease typing $1 + 4 + 9 … + n^2 = (n/6)(n+1)(2n+1)$ in Google Search.

edited Dec 15 '14 at 04:52

dustin

8,241

answered Oct 15 '14 at 03:22

user129360

175

Prove that $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$

4 Answers4