anonymous
  • anonymous
Proof help? http://imgur.com/bqFCa
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
So first step of mathematical induction is show the first case is true, so i plug in 1 for i on the left side and n on the right side:
anonymous
  • anonymous
lol that's from reddit
anonymous
  • anonymous
\[\sum_{1}^{i=1}(1)2^{(1)}=\]

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anonymous
  • anonymous
Whats from reddit?
anonymous
  • anonymous
So above, on the left i have 2
anonymous
  • anonymous
on the right....
anonymous
  • anonymous
It's the sum up to n + 1, not n
anonymous
  • anonymous
\[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]
anonymous
  • anonymous
@ strobe, so i am supposed to plug in a 2 instead of a 1?
anonymous
  • anonymous
It's the sum of the LHS subbed with 1 and 2
anonymous
  • anonymous
LHS?
anonymous
  • anonymous
left hand side
anonymous
  • anonymous
Yah
anonymous
  • anonymous
ok so the RHS is right
anonymous
  • anonymous
\[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]
anonymous
  • anonymous
so add that to the first summation and i get 10 on the LHS
anonymous
  • anonymous
so 10 = 10 is true
anonymous
  • anonymous
Is that correct?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Ok, thanks, ill see if I can finish the rest from there :)
anonymous
  • anonymous
Good luck!
anonymous
  • anonymous
after step one (show p(1) is true: step 2 is assume p(k) is true and then show p(k) implies p(k+1)
anonymous
  • anonymous
Cant quite see how i can make that happen at the moment, heres what I have:
anonymous
  • anonymous
\[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]
anonymous
  • anonymous
\[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]
anonymous
  • anonymous
RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]
anonymous
  • anonymous
So now I need to add something to p(k) to make it match p(k+1)
anonymous
  • anonymous
and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof
anonymous
  • anonymous
I wouldn't bother expanding the RHS. I would expand the LHS, then put it in terms of p(k) + something, replace p(k) with the LHS, then rearrange that result so it looks like the LHS of p(k+1)
anonymous
  • anonymous
\[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]
anonymous
  • anonymous
Hmm, i was always taught to modify add to p(k) then modify p(k)+ whateverRHS to make it the same as RHS of p(k+1)
anonymous
  • anonymous
This is so freakin complicated
anonymous
  • anonymous
"I would calculate the RHS for k + 1 and then leave it alone. Then try and change the LHS to the same form as the RHS but substituting in the the RHS of p(k) into the LHS."
anonymous
  • anonymous
So first I should calculate the right hand side of p(k+1)?
anonymous
  • anonymous
or of p(k)
anonymous
  • anonymous
p(k+1), but dont expand or simplify it, leave it as it is. This is what you're aiming for with the LHS.
anonymous
  • anonymous
the LHS of p(k+1)
anonymous
  • anonymous
ok so,
anonymous
  • anonymous
RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]
anonymous
  • anonymous
isnt much calculation to do really, just plug in
anonymous
  • anonymous
Then I want to make LHS of P(k+1) the same as that:
anonymous
  • anonymous
RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]
anonymous
  • anonymous
*LHS rather
anonymous
  • anonymous
Yup, so that equals the LHS of p(k) + something, find that something
anonymous
  • anonymous
So I should expand the LHS of p(k+1) first?
anonymous
  • anonymous
Expand so it looks like LHS of p(k) + something
anonymous
  • anonymous
I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal
anonymous
  • anonymous
I just have to show that...
anonymous
  • anonymous
Okay, so LHS p(k+1) = LHS p(k) + (k+2)2^(k+2) (you almost got it) but we know LHS = RHS of p(k), so substitute the LHS with the RHS
anonymous
  • anonymous
\[(p(k)LHS)+(k+1)2 ^{k+1}-> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]
anonymous
  • anonymous
The substitution into (p(k)LHS) is p(k)RHS
anonymous
  • anonymous
Now I need to make that = the RHS or p(k+1)?
anonymous
  • anonymous
If so, that's the method I am familiar with
anonymous
  • anonymous
Yup, you'll find it all factorises nicely
anonymous
  • anonymous
ok, I must have bliped up somewhere before then, cause the Q i asked making the sides equal was my next step from here attempted
anonymous
  • anonymous
I'll try it here
anonymous
  • anonymous
so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]
anonymous
  • anonymous
expanding everything out isnt the way to go?
anonymous
  • anonymous
yes
anonymous
  • anonymous
it is the way to go?
anonymous
  • anonymous
hmm
anonymous
  • anonymous
p(k): \[k2^{k}4+2^{k}2+4\]
anonymous
  • anonymous
Im not seeing it :/
anonymous
  • anonymous
Clearly, im totally lost
phi
  • phi
You have to be very careful with the indices with this problem. We can show the expression works for n=1. Now assume it is true for n=m. Can we show it is true for m+1?
anonymous
  • anonymous
Thats what i tried to show
anonymous
  • anonymous
Did i make an arithmetic error somewhere or is my methodology wrong or what
phi
  • phi
\[\sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)w^{m+2}\]
phi
  • phi
not sure where you went wrong. See how we start: We increased m by 1, and then break the sum into two parts.
phi
  • phi
the first sum on the right hand side "works" because we assume it works for n=m. so we can replace it with \[ m 2^{m+2} +2 \]
phi
  • phi
that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
anonymous
  • anonymous
ok
anonymous
  • anonymous
so now I need to set that equal to the right hand side of m+1
anonymous
  • anonymous
or make it equal that at least
phi
  • phi
I'm still working on it. we want the right hand side equal to what the formula says if we use n= m+1
anonymous
  • anonymous
RHS of m+1:\[(m+1)2^{m+3}+2 \]
phi
  • phi
that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
anonymous
  • anonymous
right we need to make that equal what i just typed up right?
anonymous
  • anonymous
If so, that is this: http://imgur.com/P8Tnf
phi
  • phi
yes and I believe they are.
anonymous
  • anonymous
Just with m
anonymous
  • anonymous
isnt it or am i missing something?
phi
  • phi
we have m+2 now, you have k+1
phi
  • phi
very close, but not close enough!
anonymous
  • anonymous
ok i must have messed that up somewhere
phi
  • phi
Do you see how to get the "new" version? (with m+2), it's the next term we are adding, with i= (m+2)
phi
  • phi
It's confusing because we go to n+1 instead of n, so going up one term means n+2. Anyways... can you simplify what we have?
anonymous
  • anonymous
yea, so confusing!
phi
  • phi
multiply out (m+2) 2^(m+2)
anonymous
  • anonymous
making it m made it less tho
anonymous
  • anonymous
lets see
phi
  • phi
we get \[ 2+ m 2^{m+2}+m2^{m+2} + 2^{m+3} \]
anonymous
  • anonymous
on LHS right
anonymous
  • anonymous
thats not the same tho
anonymous
  • anonymous
thats \[2+2m ^{m+2}+2^{m+3}\]
anonymous
  • anonymous
on the LHS
phi
  • phi
add the two m2^(m+2) to get m*2*2^(m+2) = m*2^(m+3)
anonymous
  • anonymous
RHS is \[2+m2^{m+3}+2^{m+3}+2\]
anonymous
  • anonymous
Ah yes
anonymous
  • anonymous
Now I see it :)
phi
  • phi
you have an extra 2 in there. But factor out 2^(m+3) to get \[ 2+ (m+1) 2^{m+3} \]
anonymous
  • anonymous
Let me try and write this out and think it out
anonymous
  • anonymous
hopefully fully see where i bliped up
phi
  • phi
ok have fun
anonymous
  • anonymous
Thanks so much
anonymous
  • anonymous
Will post here when done
anonymous
  • anonymous
http://imgur.com/BYB4E
phi
  • phi
I would tweak it a bit. Your statement showing that the formula works for p(1) is not clear. It is better to state: by definition \[ p(1)=\sum_{1}^{2} i2^i= 1\cdot2^1+2\cdot 2^2= 10\] From the formula, p(1) is \[ p(1)= 1 \cdot 2^{(1+2)}+2=10 \] therefore the formula holds for n=1 Also add a short explanation of what you are doing in part 2b. For example, The formula gives p(k+1) as .... and for the second part: By definition, p(k+1)= .... Finally, point out where you are making use of the assumption that the formula is assumed to be true for p(k). This is where you replace the summation with the results of the formula.

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