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mwmnjBest ResponseYou've already chosen the best response.2
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:
 2 years ago

neverforgetvivisteeBest ResponseYou've already chosen the best response.0
lol that's from reddit
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
\[\sum_{1}^{i=1}(1)2^{(1)}=\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
So above, on the left i have 2
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
It's the sum up to n + 1, not n
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
\[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
@ strobe, so i am supposed to plug in a 2 instead of a 1?
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
It's the sum of the LHS subbed with 1 and 2
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
\[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
so add that to the first summation and i get 10 on the LHS
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
Ok, thanks, ill see if I can finish the rest from there :)
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
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)
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
Cant quite see how i can make that happen at the moment, heres what I have:
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
\[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
\[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
So now I need to add something to p(k) to make it match p(k+1)
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
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)
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
\[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
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)
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
This is so freakin complicated
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
"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."
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
So first I should calculate the right hand side of p(k+1)?
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
p(k+1), but dont expand or simplify it, leave it as it is. This is what you're aiming for with the LHS.
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
isnt much calculation to do really, just plug in
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
Then I want to make LHS of P(k+1) the same as that:
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
Yup, so that equals the LHS of p(k) + something, find that something
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
So I should expand the LHS of p(k+1) first?
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
Expand so it looks like LHS of p(k) + something
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
I just have to show that...
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
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
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
\[(p(k)LHS)+(k+1)2 ^{k+1}> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
The substitution into (p(k)LHS) is p(k)RHS
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
Now I need to make that = the RHS or p(k+1)?
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
If so, that's the method I am familiar with
 2 years ago

strobeBest ResponseYou've already chosen the best response.0
Yup, you'll find it all factorises nicely
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
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
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
expanding everything out isnt the way to go?
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
p(k): \[k2^{k}4+2^{k}2+4\]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
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?
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
Thats what i tried to show
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
Did i make an arithmetic error somewhere or is my methodology wrong or what
 2 years ago

phiBest ResponseYou've already chosen the best response.1
\[\sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)w^{m+2}\]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
not sure where you went wrong. See how we start: We increased m by 1, and then break the sum into two parts.
 2 years ago

phiBest ResponseYou've already chosen the best response.1
the first sum on the right hand side "works" because we assume it works for n=m. so we can replace it with \[ m 2^{m+2} +2 \]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
so now I need to set that equal to the right hand side of m+1
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
or make it equal that at least
 2 years ago

phiBest ResponseYou've already chosen the best response.1
I'm still working on it. we want the right hand side equal to what the formula says if we use n= m+1
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
RHS of m+1:\[(m+1)2^{m+3}+2 \]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
right we need to make that equal what i just typed up right?
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
If so, that is this: http://imgur.com/P8Tnf
 2 years ago

phiBest ResponseYou've already chosen the best response.1
yes and I believe they are.
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
isnt it or am i missing something?
 2 years ago

phiBest ResponseYou've already chosen the best response.1
we have m+2 now, you have k+1
 2 years ago

phiBest ResponseYou've already chosen the best response.1
very close, but not close enough!
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
ok i must have messed that up somewhere
 2 years ago

phiBest ResponseYou've already chosen the best response.1
Do you see how to get the "new" version? (with m+2), it's the next term we are adding, with i= (m+2)
 2 years ago

phiBest ResponseYou've already chosen the best response.1
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?
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
making it m made it less tho
 2 years ago

phiBest ResponseYou've already chosen the best response.1
we get \[ 2+ m 2^{m+2}+m2^{m+2} + 2^{m+3} \]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
thats \[2+2m ^{m+2}+2^{m+3}\]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
add the two m2^(m+2) to get m*2*2^(m+2) = m*2^(m+3)
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
RHS is \[2+m2^{m+3}+2^{m+3}+2\]
 2 years ago

phiBest ResponseYou've already chosen the best response.1
you have an extra 2 in there. But factor out 2^(m+3) to get \[ 2+ (m+1) 2^{m+3} \]
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
Let me try and write this out and think it out
 2 years ago

mwmnjBest ResponseYou've already chosen the best response.2
hopefully fully see where i bliped up
 2 years ago

phiBest ResponseYou've already chosen the best response.1
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.
 2 years ago
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