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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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
 4 years ago
Best ResponseYou've already chosen the best response.0lol that's from reddit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{1}^{i=1}(1)2^{(1)}=\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So above, on the left i have 2

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ strobe, so i am supposed to plug in a 2 instead of a 1?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so the RHS is right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so add that to the first summation and i get 10 on the LHS

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, thanks, ill see if I can finish the rest from there :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0after 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
 4 years ago
Best ResponseYou've already chosen the best response.0Cant quite see how i can make that happen at the moment, heres what I have:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So now I need to add something to p(k) to make it match p(k+1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm, 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
 4 years ago
Best ResponseYou've already chosen the best response.0This is so freakin complicated

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"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
 4 years ago
Best ResponseYou've already chosen the best response.0So first I should calculate the right hand side of p(k+1)?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0isnt much calculation to do really, just plug in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then I want to make LHS of P(k+1) the same as that:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So I should expand the LHS of p(k+1) first?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just have to show that...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[(p(k)LHS)+(k+1)2 ^{k+1}> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The substitution into (p(k)LHS) is p(k)RHS

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now I need to make that = the RHS or p(k+1)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If so, that's the method I am familiar with

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, 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
 4 years ago
Best ResponseYou've already chosen the best response.0so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0expanding everything out isnt the way to go?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0p(k): \[k2^{k}4+2^{k}2+4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Clearly, im totally lost

phi
 4 years ago
Best ResponseYou've already chosen the best response.1You 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
 4 years ago
Best ResponseYou've already chosen the best response.0Thats what i tried to show

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Did i make an arithmetic error somewhere or is my methodology wrong or what

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

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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1the 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
 4 years ago
Best ResponseYou've already chosen the best response.1that "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
 4 years ago
Best ResponseYou've already chosen the best response.0so now I need to set that equal to the right hand side of m+1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or make it equal that at least

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0RHS of m+1:\[(m+1)2^{m+3}+2 \]

phi
 4 years ago
Best ResponseYou've already chosen the best response.1that "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
 4 years ago
Best ResponseYou've already chosen the best response.0right we need to make that equal what i just typed up right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If so, that is this: http://imgur.com/P8Tnf

phi
 4 years ago
Best ResponseYou've already chosen the best response.1yes and I believe they are.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0isnt it or am i missing something?

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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1very close, but not close enough!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok i must have messed that up somewhere

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

phi
 4 years ago
Best ResponseYou've already chosen the best response.1It'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
 4 years ago
Best ResponseYou've already chosen the best response.0making it m made it less tho

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats not the same tho

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats \[2+2m ^{m+2}+2^{m+3}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0RHS is \[2+m2^{m+3}+2^{m+3}+2\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let me try and write this out and think it out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hopefully fully see where i bliped up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Will post here when done

phi
 4 years ago
Best ResponseYou've already chosen the best response.1I 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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