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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?

no, base case is \(n=1\)

in which case you get 3, so done with the base case

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?

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

oops i meant
\[(k+1)^3+2(k+1)\]

go ahead and expand it, let me know what you get

still doing the algebra?

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

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
\[\color{blue}{k^3+2k}+\color{red}{3k^2+3k+3}\]

if the red wasn't divisible would it not be a proof?

i can't believe i found someone to walk me through that

if i can find out how to donate to you i will :)