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

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what should our first step be?

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0I have written some stuff for the problem...so..dw:1441663294530:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1your #2 is a little wrong,dw:1441663385737:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.0now add (k+ 1)^2 to the sum of k terms

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[P(k)+(k+1)^2~\color{red}{?=?}~\frac{k(k+1)(2k+1)}{6}+(k+1)^2\] 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
 one year ago
Best ResponseYou've already chosen the best response.0so k^2? before (k+1)^2 on the left side?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1P(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
 one year ago
Best ResponseYou've already chosen the best response.1bah, i put a ^2 where there shouldnt be one ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{k(k+1)(2k+1)}{6}+(k+1)^2\color{red}{\implies}\frac{(k+1)~((k+1)+1)~(2(k+1)+1)}{6}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1as 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
 one year ago
Best ResponseYou've already chosen the best response.0why didn't you do k(k+1) and turned it into k^2+k?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1because that is NOT the format that we are trying to prove ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1proof by induction is ALL about the FORMAT ... we cannot go changing the format

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we want to work the (k+1)th term INTO the format ... not the other way around.

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0how did you get rid of the (k+1)^2? on the right side?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i didnt ... i just simply showed that we need to go from one form, to the other ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1id start by getting it all over the same denominator, then parsing out the numerator

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1now, 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
 one year ago
Best ResponseYou've already chosen the best response.1err, right into left ... never was good with sides lol

Bee_see
 one year ago
Best ResponseYou've already chosen the best response.0doesn't the numerator become k(k+1)(2k+1)+6(k+1)^2 ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1now we can factor our a (k+1) to get started

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1k(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
 one year ago
Best ResponseYou've already chosen the best response.0that still doesn't get rid of that 6 on the numerator

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1:) 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
 one year ago
Best ResponseYou've already chosen the best response.1now 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
 one year ago
Best ResponseYou've already chosen the best response.0I thought we wanted to end up with (k+1)[ (k+1)+1) ] [2(k+1)+1]...aren't you missing the (k+1)?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1no, 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 ....

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1we already factored out a (k+1) ... no need to worry about it.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{(k+1)\color{red}{[k(2k+1)+6(k+1)]}}{6}\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{`\color{red}{[k(2k+1)+6(k+1)]}}{`}\] \[\frac{`\color{red}{[2k^2+7k+6]}}{`}\] \[\frac{`\color{red}{[2k^2+\color{green}{(4k+3k)}+6]}}{`}\] \[\frac{`\color{red}{[(2k^2+4k)+(3k+6)]}}{`}\] \[\frac{`\color{red}{[2k(k+2)+3(k+2)]}}{`}\] \[\frac{`\color{red}{(k+2)(2k+3)}}{`}\] \[\frac{`\color{red}{(k+2)(2k+\color{green}{(2+1)})}}{`}\] \[\frac{`\color{red}{(k+2)((2k+2)+1)}}{`}\] etc ...
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