Prove by induction that
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)/6
for every positive integer n.

- Bee_see

Prove by induction that
12 + 22 + · · · + n2 =
n(n + 1)(2n + 1)/6
for every positive integer n.

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- amistre64

what should our first step be?

- Bee_see

I have written some stuff for the problem...so..|dw:1441663294530:dw|

- Bee_see

assuming Sk is true

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## More answers

- amistre64

your #2 is a little wrong,|dw:1441663385737:dw|

- Bee_see

|dw:1441663404145:dw|

- amistre64

i tend to use P(k)
\[P(k)=12+22+...+k^2=\frac{k(k+1)(2k+1)}{6}\]
or simply
\[P(k)=\frac{k(k+1)(2k+1)}{6}\]
now the idea is to add the next term to each side ... what is the next term?

- welshfella

now add (k+ 1)^2 to the sum of k terms

- Bee_see

(k+1)^2

- Bee_see

|dw:1441663539407:dw|

- amistre64

\[P(k)+(k+1)^2~\color{red}{?=?}~\frac{k(k+1)(2k+1)}{6}+(k+1)^2\]
the idea is to get it into the same FORM, since the only thing we really know is that the FORM is good for k=1

- Bee_see

so k^2? before (k+1)^2 on the left side?

- Bee_see

no K^2*

- amistre64

P(k) is equal to all the kth term stuff, so all we need to do is add the next term to it
can we rework thru valid mathimatical procesess
\[\frac{k(k+1)(2k+1)}{6}+(k+1)^2\color{red}{\implies}\frac{(k+1)^2((k+1)+1)(2(k+1)+1)}{6}\]

- amistre64

bah, i put a ^2 where there shouldnt be one ...

- amistre64

\[\frac{k(k+1)(2k+1)}{6}+(k+1)^2\color{red}{\implies}\frac{(k+1)~((k+1)+1)~(2(k+1)+1)}{6}\]

- amistre64

as long as we can work it thru from left format, to the right format, we are good as far as proof by induction goes

- Bee_see

why didn't you do k(k+1) and turned it into k^2+k?

- amistre64

because that is NOT the format that we are trying to prove ...

- amistre64

proof by induction is ALL about the FORMAT ... we cannot go changing the format

- amistre64

we want to work the (k+1)th term INTO the format ... not the other way around.

- Bee_see

how did you get rid of the (k+1)^2? on the right side?

- amistre64

i didnt ... i just simply showed that we need to go from one form, to the other ...

- amistre64

id start by getting it all over the same denominator, then parsing out the numerator

- amistre64

now, as a guide, you can work from the left format into the right format ... but that only proves you can work it one way, and not back again ... as long as you can show valid mathematical processes to work it back again your proof is done

- amistre64

err, right into left ... never was good with sides lol

- Bee_see

doesn't the numerator become k(k+1)(2k+1)+6(k+1)^2 ?

- amistre64

yes

- amistre64

now we can factor our a (k+1) to get started

- amistre64

k(k+1)(2k+1)+6(k+1)^2
^^^ ^^^
(k+1) [(2k+1)+6(k+1)]
since we want to start it with a (k+1) we have that covered

- Bee_see

ok

- Bee_see

that still doesn't get rid of that 6 on the numerator

- amistre64

:) lets guide ourselves backwards
we want to end up with
((k+1)+1)(2(k+1)+1)
(k+2)(2k+2+1)
(k+2)(2k+3)
2k^2 +4k +3k +6
2k^2 +7k +6
.....................
expand what we got
k(2k+1)+6(k+1)
2k^2 +k +6k +6
2k^2 +7k +6
they match up, so lets just work the process backwards

- amistre64

now that we have a path to work from ...
we can expand everything to here:
2k^2 +7k +6
split the 7k into 3k + 4k
2k^2 +3k + 4k +6
and facotr

- Bee_see

I thought we wanted to end up with (k+1)[ (k+1)+1) ] [2(k+1)+1]...aren't you missing the (k+1)?

- amistre64

no, im not missing it, im just ignoring it at the moment so i dont have to keep writing it down when we no longer have to focus on it ....

- Bee_see

ok

- amistre64

we already factored out a (k+1) ... no need to worry about it.

- amistre64

\[\frac{(k+1)\color{red}{[k(2k+1)+6(k+1)]}}{6}\]

- amistre64

\[\frac{`\color{red}{[k(2k+1)+6(k+1)]}}{`}\]
\[\frac{`\color{red}{[2k^2+7k+6]}}{`}\]
\[\frac{`\color{red}{[2k^2+\color{green}{(4k+3k)}+6]}}{`}\]
\[\frac{`\color{red}{[(2k^2+4k)+(3k+6)]}}{`}\]
\[\frac{`\color{red}{[2k(k+2)+3(k+2)]}}{`}\]
\[\frac{`\color{red}{(k+2)(2k+3)}}{`}\]
\[\frac{`\color{red}{(k+2)(2k+\color{green}{(2+1)})}}{`}\]
\[\frac{`\color{red}{(k+2)((2k+2)+1)}}{`}\]
etc ...

- Bee_see

ok

- Bee_see

I get it now.

- amistre64

good luck then :)

- Bee_see

Thanks for the help.

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