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Mathematics 66 Online
OpenStudy (anonymous):

I can't quite get the answer... PLEASE HELP!!! For the given statement Pn, write the statements P1, Pk, and Pk+1. 3. 2 + 4 + 8 + . . . + 2n = 2(n+1) - 2 P1 2(n+1)-2 2(1+1)-2 4-2=2 (first term) Pk 2+4+8+…+2k=2(k+1)-2 Pk+1 2+4+8+…+2k+2(k+1)=2(k+1+1)-2 2+4+8+…+2k+2(k+1)=2(k+2)-2 2(k+1)-2+2(k+1)=2(k+2)-2 2k+2-2+2k+2=2(k+2)-2 4k+2=2(k+1)-2

OpenStudy (anonymous):

we can do this, give me a minute to get to a real computer

OpenStudy (anonymous):

@satellite73 : may u check my question when u finish this question ? Ty

OpenStudy (anonymous):

okay! :)

OpenStudy (anonymous):

wow it took a long time to load

OpenStudy (anonymous):

first of all, \[2 + 4 + 8 + . . . + 2n = 2(n+1) - 2\] is not correct

OpenStudy (anonymous):

\[2(n+1)-2=2,+2-2=2n\] which is only the last term

OpenStudy (anonymous):

it should be \[2+4+6+8+...+2n=n^2+n\] is that what you have?

OpenStudy (anonymous):

@Eyad sure, what would you like to me check @madixy it looks like you are doing a proof "by induction" so we need the right formula

OpenStudy (anonymous):

http://openstudy.com/study#/updates/513e8e0ae4b029b0182c2de0 This one ,if u dont mind

OpenStudy (anonymous):

Sry madixy for ruining your question.

OpenStudy (anonymous):

oh sorry i was waiting and forgot to check if you were back... okay I'm going to read what you wrote real quick @satellite73 and you're not ruining it it's fine! :) @Eyad

OpenStudy (anonymous):

so pk is wrong?

OpenStudy (anonymous):

the sum of the first \(n\) even numbers is \(n^2+n\)

OpenStudy (anonymous):

not \(2(n+1)-2\)which is just \(2n\)

OpenStudy (anonymous):

in other words, it is \[2+4+6+...+2n=n^2+n\]

OpenStudy (anonymous):

\(P_1\) would be \(2=1^2+1\) which is true

OpenStudy (anonymous):

\(P_k\) would be \[2+4+6+...+2k=k^2+k\]

OpenStudy (anonymous):

Wait but it says 2(n+1)-2 in the question...?

OpenStudy (anonymous):

and \(P_{k+1}\) would be \[2+4+6+...+2k +2(k+1)=(k+1)^2+(k+1)\]

OpenStudy (anonymous):

can you post or attach the exact question?

OpenStudy (anonymous):

we see from elementary algebra that \(2(n+1)-2=2n\) so it is a bit odd to have that plus, that is not what the sum is

OpenStudy (anonymous):

OpenStudy (anonymous):

Oh it would be 2n... Gotcha

OpenStudy (anonymous):

3. 2 + 4 + 8 + . . . + 2n = 2(n+1) – 2 2 + 4 + 8 + . . . + 2n = 2(n) P1 2(1) =2 (first term) Pk 2+4+8+…+2k=2(k) Pk+1 2+4+8+…+2k+2(k+1)=2(k+1) still having trouble @satellite73

OpenStudy (anonymous):

okay that is what it says, so that is what it says, i cannot argue it is, however, not true

OpenStudy (anonymous):

we can still write it \[P_1\] would say \[2(1)=2(1+1)-2\] which is in fact true

OpenStudy (anonymous):

\(P_k\) would be \[2+4+6+...+2k=2(k+1)-2\]

OpenStudy (anonymous):

which is false

OpenStudy (anonymous):

since the term on the right is \(2k\) and also the very last term on the left is \(2k\) so there is no way that they are equal

OpenStudy (anonymous):

and \(P_{k+1}\) would be \[2+4+6+...+2k+2(k+1)=2(k+1+1)-2\] which is also false so this is a bunch of nonsense, but it is what you were given so i cannot argue with the fact that you were given this question

OpenStudy (anonymous):

no wonder it is confusing, it is totally wrong but i suppose that is okay, so long as you didn't have to prove it at the risk of repeating myself, the correct formula would be \[2+4+6+...+2n=n^2+n\]

OpenStudy (anonymous):

which we could prove by induction if you are so inclined, or else we could forget it

OpenStudy (anonymous):

@satellite73 : Are u a math teacher or something ? or u its just your interest ?

OpenStudy (anonymous):

Okay can you maybe help me go through pk-1 please?

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

\(P_{k-1}\) means replace \(k\) by \(k-1\)

OpenStudy (anonymous):

\[2+4+6+...+2(k-1)=2(k-1+1)-2\] would be what you would write, but it is totally annoying because it is totally wrong!!

OpenStudy (anonymous):

who wrote this question??

OpenStudy (anonymous):

virtual school lol

OpenStudy (anonymous):

do I have to factor it out?

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

i don't know if you have to multiply it out. the whole thing is ridiculous, sorry

OpenStudy (anonymous):

if you do, you get \(2k-2\)

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

Oh my computer is messing up sorry! Thanks for your help!!

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