using induction, prove that for all integers n > 0, n3 + 2n is divisible by 3. please help.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

using induction, prove that for all integers n > 0, n3 + 2n is divisible by 3. please help.

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

first of all maybe we can see why it is true without induction
or just do it by induction either way have you ever done a proof by induction?
i'm supposed to solve it using induction. it's obvious that it's true. even if i plug in 0 it works. but, i don't understand understand how to set this up. i know i need a base case. would that be n=0?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

no, base case is \(n=1\)
in which case you get 3, so done with the base case
next you assume it is true if \(n=k\) in other words you say "we assume \(k^3+2k\) is divisible by \(3\)" and using that, prove it is true if \(n=k+1\)
most of it now is algebra from here on in replace \(n\) by \(k+1\) and separate out a part of the result that looks like \(k^3+2k\) and the rest will be obviously divisible by \(3\)
one sec let me try to absorb what you said. thank you for helping me. i will reply in a min with questions
take your time, do the algebra
what is the significance of substituting n for k?
you substitute \(k\) for \(n\) to say what you get to assume is true
not \(n\) for \(k\)
ok, so by saying k we are just saying we assume it's true. this signifies induction?
and if k+1 is true then the proof is complete?
yes that is the "induction hypothesis" assume it is true if \(n=k\) then you use that to prove it is true if \(n=k+1\)
some people cheat and assume it is true for \(n\) then show it is true for \(n+1\) but really we should change the variable
are you working on the algebra?
i am trying to work on the algebra. this is what i have to solve: (k+1)^3+2(k+1)=3
oh no not \(=3\) no one says it is equal to 3
just take \[(k+1)^2+2(k+1)\] and expand it, then take out the part that looks like \(k^3+2k\) and separate it from the rest
oops i meant \[(k+1)^3+2(k+1)\]
go ahead and expand it, let me know what you get
still doing the algebra?
i am.. i feel retarded. thanks again for helping :\ i am trying to figure out what else i do after (k+1)((k+1)^2)+2(k+1)
multiply it all out there is no way around it
do you know how to cube \(k+1\) easily?
in general \[(a+b)^3=a^2+3a^2b+3ab^2+b^2\] in your case \(a=k,, b=1\)
and of course \(2(k+1)=2k+2\)
okay. i have to do (k+1)(k+1)(k+1) and multiply all of it.
yes, but i showed you how to do it in one step
yeah i got it. sorry about that
take your time, let me know what you get when you do all this algebra then we will be done in two steps after
what did you mean by separate it from the rest? it's all multiplied out. k^2+3k^2(1)+3k(1)^2+1^2 + 2k+1 but there is no right side
ok you should have \[k^3+3k^2+3k+1+2k+2\]
you forgot to distribute with \(2(k+1)\) i think
and also you wrote \(1^2\) but i am sure you didn't mean to
now only two steps separate out the \(k^3+2k\) part, that we get to assume is divisible by 3
you have \[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\]
the blue part is divisible by 3 "by induction" we got to assume that the red part is divisible by 3 because... well because it is you can factor a 3 out of it if you want to be formal
if the red wasn't divisible would it not be a proof?
yeah you need that part to be divisible by 3 the blue part is the induction hypotheses the red part is what you have to show is divisible by 3, but there is not much to show since each term has a 3 in it
\[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\] \[\overbrace{\color{blue}{k^3+2k}}^{\text{ div by 3 by induction}}+\overbrace{\color{red}{3(k^2+k+1)}}^{\text{ has a factor of 3}}\]
and now we are done hope this helped clarify how to do a proof by induction they all pretty much work this way
i can't believe i found someone to walk me through that
yeah it is usually a thankless task because most people have no idea about what it means so you have to explain every step of the way, and they get frustrated really it is something you have to see done a few times to understand the logic glad it helped
if i can find out how to donate to you i will :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question