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mwmnj
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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:
neverforgetvivistee
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lol that's from reddit
mwmnj
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\[\sum_{1}^{i=1}(1)2^{(1)}=\]
mwmnj
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Whats from reddit?
mwmnj
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So above, on the left i have 2
mwmnj
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on the right....
strobe
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It's the sum up to n + 1, not n
mwmnj
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\[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]
mwmnj
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@ strobe, so i am supposed to plug in a 2 instead of a 1?
strobe
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It's the sum of the LHS subbed with 1 and 2
mwmnj
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LHS?
mwmnj
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left hand side
strobe
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Yah
mwmnj
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ok so the RHS is right
mwmnj
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\[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]
mwmnj
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so add that to the first summation and i get 10 on the LHS
mwmnj
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so 10 = 10 is true
mwmnj
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Is that correct?
strobe
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Yes
mwmnj
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Ok, thanks, ill see if I can finish the rest from there :)
strobe
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Good luck!
mwmnj
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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)
mwmnj
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Cant quite see how i can make that happen at the moment, heres what I have:
mwmnj
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\[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]
mwmnj
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\[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]
mwmnj
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RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]
mwmnj
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So now I need to add something to p(k) to make it match p(k+1)
mwmnj
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and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof
strobe
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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)
mwmnj
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\[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]
mwmnj
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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)
mwmnj
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This is so freakin complicated
mwmnj
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"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."
mwmnj
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So first I should calculate the right hand side of p(k+1)?
mwmnj
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or of p(k)
strobe
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p(k+1), but dont expand or simplify it, leave it as it is. This is what you're aiming for with the LHS.
mwmnj
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the LHS of p(k+1)
mwmnj
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ok so,
mwmnj
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RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]
mwmnj
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isnt much calculation to do really, just plug in
mwmnj
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Then I want to make LHS of P(k+1) the same as that:
mwmnj
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RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]
mwmnj
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*LHS rather
strobe
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Yup, so that equals the LHS of p(k) + something, find that something
mwmnj
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So I should expand the LHS of p(k+1) first?
strobe
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Expand so it looks like LHS of p(k) + something
mwmnj
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I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal
mwmnj
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I just have to show that...
strobe
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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
mwmnj
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\[(p(k)LHS)+(k+1)2 ^{k+1}-> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]
mwmnj
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The substitution into (p(k)LHS) is p(k)RHS
mwmnj
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Now I need to make that = the RHS or p(k+1)?
mwmnj
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If so, that's the method I am familiar with
strobe
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Yup, you'll find it all factorises nicely
mwmnj
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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
mwmnj
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I'll try it here
mwmnj
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so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]
mwmnj
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expanding everything out isnt the way to go?
mwmnj
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it is the way to go?
mwmnj
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hmm
mwmnj
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p(k): \[k2^{k}4+2^{k}2+4\]
mwmnj
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Im not seeing it :/
mwmnj
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Clearly, im totally lost
phi
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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?
mwmnj
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Thats what i tried to show
mwmnj
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Did i make an arithmetic error somewhere or is my methodology wrong or what
phi
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\[\sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)w^{m+2}\]
phi
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not sure where you went wrong.
See how we start: We increased m by 1, and then break the sum into two parts.
phi
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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
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that "w" in the first formula is a 2 !
\[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\]
\[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
mwmnj
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ok
mwmnj
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so now I need to set that equal to the right hand side of m+1
mwmnj
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or make it equal that at least
phi
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I'm still working on it. we want the right hand side equal to what the formula says if we use n= m+1
mwmnj
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RHS of m+1:\[(m+1)2^{m+3}+2 \]
phi
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that "w" in the first formula is a 2 !
\[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\]
\[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
mwmnj
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right we need to make that equal what i just typed up right?
phi
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yes
and I believe they are.
mwmnj
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Just with m
mwmnj
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isnt it or am i missing something?
phi
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we have m+2 now, you have k+1
phi
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very close, but not close enough!
mwmnj
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ok i must have messed that up somewhere
phi
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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
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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?
mwmnj
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yea, so confusing!
phi
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multiply out (m+2) 2^(m+2)
mwmnj
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making it m made it less tho
mwmnj
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lets see
phi
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we get
\[ 2+ m 2^{m+2}+m2^{m+2} + 2^{m+3} \]
mwmnj
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on LHS right
mwmnj
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thats not the same tho
mwmnj
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thats \[2+2m ^{m+2}+2^{m+3}\]
mwmnj
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on the LHS
phi
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add the two m2^(m+2) to get m*2*2^(m+2) = m*2^(m+3)
mwmnj
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RHS is \[2+m2^{m+3}+2^{m+3}+2\]
mwmnj
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Ah yes
mwmnj
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Now I see it :)
phi
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you have an extra 2 in there. But factor out 2^(m+3) to get
\[ 2+ (m+1) 2^{m+3} \]
mwmnj
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Let me try and write this out and think it out
mwmnj
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hopefully fully see where i bliped up
phi
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ok have fun
mwmnj
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Thanks so much
mwmnj
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Will post here when done
phi
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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.