Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

- anonymous

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- schrodinger

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- anonymous

|dw:1378180823335:dw|

- zzr0ck3r

its not true for n = 1 on the left we get 24 and on the right we get 4*5*15/6 = 50

- anonymous

May you elaborate on why it is not true and how, because I don't know how to do this at all?

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## More answers

- zzr0ck3r

we are trying to prove that the left hand side equals the right hand side for all positive integers, the way we do this is through mathematical induction.
the first step, if we are trying to prove it for 1,2,3,..... is to prove it is true for n = 1
when n = 1 on the left hand side we have 4*1(4*1+2) = 4*6 = 24
when n = 1 on the right hand side we have 4*5*15/6 = 50
so since the goal is to show that it is true for all positive integers, we have already showed that is not true
because its not even true for n = 1

- zzr0ck3r

maybe if you have another example of one that we can prove is true, you might understand it better

- anonymous

Hold on.

- zzr0ck3r

the left hand side o your problem is the same thing as this\[\sum_{k=1}^{n}(4k(4k+2))\]

- zzr0ck3r

so you are trying to prove
\[\sum_{k=1}^{n}(4k(4k+2))=\frac{4(4n+1)(8n+7)}{6}\]

- anonymous

|dw:1378181422563:dw|

- zzr0ck3r

ok think this one will work

- zzr0ck3r

\[\sum_{k=1}^{n}(3n-2)^2=\frac{n(6n^2-3n-1)}{2}\]
we will show that it is true for n = 1
when n=1 we have
\[(3*1-2)^2=1\\and\\\frac{1(6-3-1)}{2}=1\]
so we know its true for n = 1
you with me so far?

- anonymous

Yes

- zzr0ck3r

so our inductive hypothesis is \[\sum_{k=1}^{n}(3n-2)^2=\frac{n(6n^2-3n-1)}{2}\]for some n
now we must show it is true for n+1\[\sum_{k=1}^{n+1}(3k-2)^2=\frac{(n+1)(6(n+1)^2-3(n+1)-1)}{2}\]start with the left hand side\[\sum_{k=1}^{n+1}(3k-2)^2=(3(n+1)-2)^2+\sum_{k=1}^{n}(3k-2)^2)=(3n+1)^2+\sum_{k=1}^{n}(3k-2)^2)\]
say when

- zzr0ck3r

by inductive hypothisis
\[(3n+1)^2+[\sum_{k=1}^{n}(3k-2)^2]=(3n+1)^2+[\frac{n(6n^2-3n-1)}{2}]\\=9n^2+6n+1+\frac{6n^3-3n^2-n}{2}=\frac{18n^2+12n+2+6n^3-3n^2-n}{2}=\\\frac{6n^3+15n^2+11n+2}{2}\]
now we need to show that is the same thing as this\[\frac{(n+1)(6(n+1)^2-3(n+1)-1)}2=\frac{(n+1)(6(n^2+2n+1)-3n-3-1)}{2}=\\\frac{(n+1)(6n^2+12n+6-3n-4)}{2}=\\\frac{(n+1)(6n^2+9n+2)}{2}=\frac{6n^3+9n^2+2n+6n^2+9n+2}{2}=\\\frac{6n^3+15n^2+11n+2}{2}\]

- zzr0ck3r

thus by induction the thing is proved:)

- anonymous

I hope you do not mind me questioning you, nonetheless are you a Professor? Honestly, you are brilliant ! I would not have known where to begin if it was not for you. I actually do comprehend the process of proving the mathematical induction, now.

- zzr0ck3r

nah I am a student.I have just seen things like this a few times:) after you do it once, you can do it again.

- zzr0ck3r

the trick with these that many people dont use, is the fact that I wrote it in sumation notation, then when we need to evaluate (n+1) we just move that off to the side, and plug in our hypotheses.

- zzr0ck3r

I know that it is odd that we are calling the hypotheses the same thing that we are trying to prove, but here is the difference
we are trying to prove that it is true for ALL n in N
our hypotheses says it is true for SOME n in N

- zzr0ck3r

if we wanted prove something was true for all positive integers >= 5 we would first start and prove it is true for n = 5 then go through the n+1 process and when we are done we know that it is true for all integers >=5

- anonymous

I'm usually great with these equations, it's just that the two previous equations were the first I've ever worked on. However, I do acknowledge what you mean.

- anonymous

Do you mind aiding me with just one more, then I'm done for the night. :)

- zzr0ck3r

I just started a movie with my wife, shes yelling at me to hurry and these take a while. So post here and ill do it when we are done and you can look in the morning

- anonymous

I do want to Thank You, for actually taking your time to elaborate every inch of the equation for me. I am honestly grateful.

- zzr0ck3r

yeah post a few to make sure that we can prove one, and ill make a nice write up...with explinantion

- zzr0ck3r

np

- anonymous

Well, this one is slightly different, but it might be a bit simple for you.

- anonymous

For the given statement Pn, write the statements P1, Pk, and Pk+1.
2 + 4 + 6 + . . . + 2n = n(n+1)

- zzr0ck3r

p1
2*1=1(1+1)
Pk
2k=k(k+1)
pk+1
2(k+1)=(k+1)((k+1)+1)

- anonymous

Thank You !

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