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Mathematics
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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:
lol that's from reddit
\[\sum_{1}^{i=1}(1)2^{(1)}=\]

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Other answers:

Whats from reddit?
So above, on the left i have 2
on the right....
It's the sum up to n + 1, not n
\[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]
@ strobe, so i am supposed to plug in a 2 instead of a 1?
It's the sum of the LHS subbed with 1 and 2
LHS?
left hand side
Yah
ok so the RHS is right
\[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]
so add that to the first summation and i get 10 on the LHS
so 10 = 10 is true
Is that correct?
Yes
Ok, thanks, ill see if I can finish the rest from there :)
Good luck!
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)
Cant quite see how i can make that happen at the moment, heres what I have:
\[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]
\[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]
RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]
So now I need to add something to p(k) to make it match p(k+1)
and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof
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)
\[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]
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)
This is so freakin complicated
"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."
So first I should calculate the right hand side of p(k+1)?
or of p(k)
p(k+1), but dont expand or simplify it, leave it as it is. This is what you're aiming for with the LHS.
the LHS of p(k+1)
ok so,
RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]
isnt much calculation to do really, just plug in
Then I want to make LHS of P(k+1) the same as that:
RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]
*LHS rather
Yup, so that equals the LHS of p(k) + something, find that something
So I should expand the LHS of p(k+1) first?
Expand so it looks like LHS of p(k) + something
I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal
I just have to show that...
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
\[(p(k)LHS)+(k+1)2 ^{k+1}-> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]
The substitution into (p(k)LHS) is p(k)RHS
Now I need to make that = the RHS or p(k+1)?
If so, that's the method I am familiar with
Yup, you'll find it all factorises nicely
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
I'll try it here
so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]
expanding everything out isnt the way to go?
yes
it is the way to go?
hmm
p(k): \[k2^{k}4+2^{k}2+4\]
Im not seeing it :/
Clearly, im totally lost
  • 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?
Thats what i tried to show
Did i make an arithmetic error somewhere or is my methodology wrong or what
  • phi
\[\sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)w^{m+2}\]
  • 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
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
that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
ok
so now I need to set that equal to the right hand side of m+1
or make it equal that at least
  • 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
RHS of m+1:\[(m+1)2^{m+3}+2 \]
  • phi
that "w" in the first formula is a 2 ! \[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\] \[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]
right we need to make that equal what i just typed up right?
If so, that is this: http://imgur.com/P8Tnf
  • phi
yes and I believe they are.
Just with m
isnt it or am i missing something?
  • phi
we have m+2 now, you have k+1
  • phi
very close, but not close enough!
ok i must have messed that up somewhere
  • 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
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?
yea, so confusing!
  • phi
multiply out (m+2) 2^(m+2)
making it m made it less tho
lets see
  • phi
we get \[ 2+ m 2^{m+2}+m2^{m+2} + 2^{m+3} \]
on LHS right
thats not the same tho
thats \[2+2m ^{m+2}+2^{m+3}\]
on the LHS
  • phi
add the two m2^(m+2) to get m*2*2^(m+2) = m*2^(m+3)
RHS is \[2+m2^{m+3}+2^{m+3}+2\]
Ah yes
Now I see it :)
  • phi
you have an extra 2 in there. But factor out 2^(m+3) to get \[ 2+ (m+1) 2^{m+3} \]
Let me try and write this out and think it out
hopefully fully see where i bliped up
  • phi
ok have fun
Thanks so much
Will post here when done
http://imgur.com/BYB4E
  • 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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