Question

Prove that:
(sigma from k=1 to n) k^2 = (n(n+1)(2n+1))/6

Help please!

Answer 1

This is a fun one. Do you know how to use mathematical induction?

Answer 2

Kinda, I am learning to

Answer 3

This is a project for a class, but I am new to writing to proofs so I am trying hard but I am stuck now! haha

Answer 4

That's fine. Learning to write proofs can sometimes be hard. Basically induction works by making a chain of logic (theorem is true for case 1, which shows it is true for case 2, which shows it is true for case 3, etc.). So first, you need to prove the base case for \(k=1
\) just by writing it out.

Answer 5

correct. I think i got that one...i think.

Answer 6

Once you've done that, you need to write "assume theorem is true for case \(k\)", which you're allowed to do because you've shown it's true for the base case. Then using that fact, you need to prove it for case \(k+1\).

Answer 7

for case k+1 or n+1?

Answer 8

Either n+1 or k+1, it doesn't really matter as long as you're consistent.

Answer 9

so, is that p(2) that i need to solve for now?

Answer 10

No, you essentially need to show that p(n+1)=p(n)+(n+1)^2. You can use \(\displaystyle p(n)=\frac{n(n+1)(2n+1)}{6}\) to help show this.

Answer 11

ok....so where do i got from that step?

Answer 12

If you've successfully proved the theorem for case n+1, all you need next is "By induction, theorem is true" and you're done!

Answer 13

lol ok i see...i see what its because of induction its true. but i dont know how to show my work on that step

Answer 14

Does it matter that it does not work for n=2?

Answer 15

It should look something like this:
\[\begin{align}
\sum_{k=1}^{n+1} k^2 &= \sum_{k=1}^n k^2 + (n+1)^2\\
&=\frac{n(n+1)(2n+1)}{6}+n^2+2n+1\\
&=\frac{2n^3+3n^2+n+6n^2+12n+6}{6}\\
&=\frac{2n^3+9n^2+13n+6}{6}\\
&=\frac{(n+1)(n+2)(2n+3)}{6}\\
&=\frac{(n+1)(n+2)(2(n+1)+1)}{6}
\end{align}\]

Answer 16

thank u so much! im gonna keep looking at it. i appreciate ur guidance!

Answer 17

:p

Answer 18

proof without words
http://a7.sphotos.ak.fbcdn.net/hphotos-ak-ash3/524120_10151990321110319_1804263112_n.jpg