Question

Proof help?
http://imgur.com/bqFCa

Answer 1

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:

Answer 2

lol that's from reddit

Answer 3

\[\sum_{1}^{i=1}(1)2^{(1)}=\]

Answer 4

Whats from reddit?

Answer 5

So above, on the left i have 2

Answer 6

on the right....

Answer 7

It's the sum up to n + 1, not n

Answer 8

\[(1)2^{((1)+2)}+2 = 2^{3}+2 = 10\]

Answer 9

@ strobe, so i am supposed to plug in a 2 instead of a 1?

Answer 10

It's the sum of the LHS subbed with 1 and 2

Answer 11

LHS?

Answer 12

left hand side

Answer 13

Yah

Answer 14

ok so the RHS is right

Answer 15

\[\sum_{i = 2}^{2} = 2(2)^{2} = 8\]

Answer 16

so add that to the first summation and i get 10 on the LHS

Answer 17

so 10 = 10 is true

Answer 18

Is that correct?

Answer 19

Yes

Answer 20

Ok, thanks, ill see if I can finish the rest from there :)

Answer 21

Good luck!

Answer 22

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)

Answer 23

Cant quite see how i can make that happen at the moment, heres what I have:

Answer 24

\[p(k) : \sum_{i=1}^{k+1}i2^{i} = k2^{k+2}+2\]

Answer 25

\[p(k+1): \sum_{i=1}^{k+2}i2^{i}=(k+1)2^{k+3}+2\]

Answer 26

RHS simplifies to \[k2^{k+3}+2^{k+3}+2\]

Answer 27

So now I need to add something to p(k) to make it match p(k+1)

Answer 28

and if after adding that to p(k) the RHS of p(k) = the RHS of p(k+1), that is sufficient proof

Answer 29

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)

Answer 30

\[p(k)+(k+1)2^{k+1}: \sum_{i=1}^{k+1}i2^{i}+(k+1)2^{k+1}=k2^{k+2}+2+(k+1)2^{k+1}\]

Answer 31

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)

Answer 32

This is so freakin complicated

Answer 33

"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."

Answer 34

So first I should calculate the right hand side of p(k+1)?

Answer 35

or of p(k)

Answer 36

p(k+1), but dont expand or simplify it, leave it as it is. This is what you're aiming for with the LHS.

Answer 37

the LHS of p(k+1)

Answer 38

ok so,

Answer 39

RHS of p(k+1): \[(k+1)2^{(k+1)+2}+2\]

Answer 40

isnt much calculation to do really, just plug in

Answer 41

Then I want to make LHS of P(k+1) the same as that:

Answer 42

RHS of p(k+1): \[\sum_{i=1}^{(k+1)+1}i2^{i}\]

Answer 43

*LHS rather

Answer 44

Yup, so that equals the LHS of p(k) + something, find that something

Answer 45

So I should expand the LHS of p(k+1) first?

Answer 46

Expand so it looks like LHS of p(k) + something

Answer 47

I would need to add (k+1)2^(k+1) to LHS of p(k) for them to be equal

Answer 48

I just have to show that...

Answer 49

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

Answer 50

\[(p(k)LHS)+(k+1)2 ^{k+1}-> (k2^{k+2}+2)+(k+1)2 ^{k+1}\]

Answer 51

The substitution into (p(k)LHS) is p(k)RHS

Answer 52

Now I need to make that = the RHS or p(k+1)?

Answer 53

If so, that's the method I am familiar with

Answer 54

Yup, you'll find it all factorises nicely

Answer 55

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

Answer 56

I'll try it here

Answer 57

so im trying to make \[(k2^{k+2}+2)+(k+1)2^{k+1} = (k+1)2^{(k+1)+2}+2 \]

Answer 58

expanding everything out isnt the way to go?

Answer 59

yes

Answer 60

it is the way to go?

Answer 61

hmm

Answer 62

p(k): \[k2^{k}4+2^{k}2+4\]

Answer 63

Im not seeing it :/

Answer 64

Clearly, im totally lost

Answer 65

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?

Answer 66

Thats what i tried to show

Answer 67

Did i make an arithmetic error somewhere or is my methodology wrong or what

Answer 68

\[\sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)w^{m+2}\]

Answer 69

not sure where you went wrong.
See how we start: We increased m by 1, and then break the sum into two parts.

Answer 70

the first sum on the right hand side "works" because we assume it works for n=m.
so we can replace it with
\[ m 2^{m+2} +2 \]

Answer 71

that "w" in the first formula is a 2 !
\[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\]
\[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]

Answer 72

ok

Answer 73

so now I need to set that equal to the right hand side of m+1

Answer 74

or make it equal that at least

Answer 75

I'm still working on it. we want the right hand side equal to what the formula says if we use n= m+1

Answer 76

RHS of m+1:\[(m+1)2^{m+3}+2 \]

Answer 77

that "w" in the first formula is a 2 !
\[ \sum_{i=1}^{m+2}i2^i= \sum_{i=1}^{m+1}i2^i + (m+2)2^{m+2}\]
\[ = 2+ m 2^{m+2} +(m+2)2^{m+2} \]

Answer 78

right we need to make that equal what i just typed up right?

Answer 79

If so, that is this: http://imgur.com/P8Tnf

Answer 80

yes
and I believe they are.

Answer 81

Just with m

Answer 82

isnt it or am i missing something?

Answer 83

we have m+2 now, you have k+1

Answer 84

very close, but not close enough!

Answer 85

ok i must have messed that up somewhere

Answer 86

Do you see how to get the "new" version? (with m+2), it's the next term we are adding, with i= (m+2)

Answer 87

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?

Answer 88

yea, so confusing!

Answer 89

multiply out (m+2) 2^(m+2)

Answer 90

making it m made it less tho

Answer 91

lets see

Answer 92

we get
\[ 2+ m 2^{m+2}+m2^{m+2} + 2^{m+3} \]

Answer 93

on LHS right

Answer 94

thats not the same tho

Answer 95

thats \[2+2m ^{m+2}+2^{m+3}\]

Answer 96

on the LHS

Answer 97

add the two m2^(m+2) to get m*2*2^(m+2) = m*2^(m+3)

Answer 98

RHS is \[2+m2^{m+3}+2^{m+3}+2\]