Bee_see
  • Bee_see
Prove by induction that 12 + 22 + · · · + n2 = n(n + 1)(2n + 1)/6 for every positive integer n.
Discrete Math
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
what should our first step be?
Bee_see
  • Bee_see
I have written some stuff for the problem...so..|dw:1441663294530:dw|
Bee_see
  • Bee_see
assuming Sk is true

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More answers

amistre64
  • amistre64
your #2 is a little wrong,|dw:1441663385737:dw|
Bee_see
  • Bee_see
|dw:1441663404145:dw|
amistre64
  • 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
  • welshfella
now add (k+ 1)^2 to the sum of k terms
Bee_see
  • Bee_see
(k+1)^2
Bee_see
  • Bee_see
|dw:1441663539407:dw|
amistre64
  • 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
  • Bee_see
so k^2? before (k+1)^2 on the left side?
Bee_see
  • Bee_see
no K^2*
amistre64
  • 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
  • amistre64
bah, i put a ^2 where there shouldnt be one ...
amistre64
  • 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
  • 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
  • Bee_see
why didn't you do k(k+1) and turned it into k^2+k?
amistre64
  • amistre64
because that is NOT the format that we are trying to prove ...
amistre64
  • amistre64
proof by induction is ALL about the FORMAT ... we cannot go changing the format
amistre64
  • amistre64
we want to work the (k+1)th term INTO the format ... not the other way around.
Bee_see
  • Bee_see
how did you get rid of the (k+1)^2? on the right side?
amistre64
  • amistre64
i didnt ... i just simply showed that we need to go from one form, to the other ...
amistre64
  • amistre64
id start by getting it all over the same denominator, then parsing out the numerator
amistre64
  • 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
  • amistre64
err, right into left ... never was good with sides lol
Bee_see
  • Bee_see
doesn't the numerator become k(k+1)(2k+1)+6(k+1)^2 ?
amistre64
  • amistre64
yes
amistre64
  • amistre64
now we can factor our a (k+1) to get started
amistre64
  • 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
  • Bee_see
ok
Bee_see
  • Bee_see
that still doesn't get rid of that 6 on the numerator
amistre64
  • 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
  • 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
  • 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
  • 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
  • Bee_see
ok
amistre64
  • amistre64
we already factored out a (k+1) ... no need to worry about it.
amistre64
  • amistre64
\[\frac{(k+1)\color{red}{[k(2k+1)+6(k+1)]}}{6}\]
amistre64
  • 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
  • Bee_see
ok
Bee_see
  • Bee_see
I get it now.
amistre64
  • amistre64
good luck then :)
Bee_see
  • Bee_see
Thanks for the help.

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