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
we can do this, give me a minute to get to a real computer
@satellite73 : may u check my question when u finish this question ? Ty
okay! :)
wow it took a long time to load
first of all, \[2 + 4 + 8 + . . . + 2n = 2(n+1) - 2\] is not correct
\[2(n+1)-2=2,+2-2=2n\] which is only the last term
it should be \[2+4+6+8+...+2n=n^2+n\] is that what you have?
@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
http://openstudy.com/study#/updates/513e8e0ae4b029b0182c2de0 This one ,if u dont mind
Sry madixy for ruining your question.
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
so pk is wrong?
the sum of the first \(n\) even numbers is \(n^2+n\)
not \(2(n+1)-2\)which is just \(2n\)
in other words, it is \[2+4+6+...+2n=n^2+n\]
\(P_1\) would be \(2=1^2+1\) which is true
\(P_k\) would be \[2+4+6+...+2k=k^2+k\]
Wait but it says 2(n+1)-2 in the question...?
and \(P_{k+1}\) would be \[2+4+6+...+2k +2(k+1)=(k+1)^2+(k+1)\]
can you post or attach the exact question?
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
Oh it would be 2n... Gotcha
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
okay that is what it says, so that is what it says, i cannot argue it is, however, not true
we can still write it \[P_1\] would say \[2(1)=2(1+1)-2\] which is in fact true
\(P_k\) would be \[2+4+6+...+2k=2(k+1)-2\]
which is false
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
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
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 \[2+4+6+...+2n=n^2+n\]
which we could prove by induction if you are so inclined, or else we could forget it
@satellite73 : Are u a math teacher or something ? or u its just your interest ?
Okay can you maybe help me go through pk-1 please?
@satellite73
\(P_{k-1}\) means replace \(k\) by \(k-1\)
\[2+4+6+...+2(k-1)=2(k-1+1)-2\] would be what you would write, but it is totally annoying because it is totally wrong!!
who wrote this question??
virtual school lol
do I have to factor it out?
@satellite73
i don't know if you have to multiply it out. the whole thing is ridiculous, sorry
if you do, you get \(2k-2\)
@satellite73
Oh my computer is messing up sorry! Thanks for your help!!
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