"Harder" induction (1 Viewer)

[ezzy85]

i came across a sheet with harder induction questions. Ive got plenty of questions, but I thought Id post just 3 and then see how I go. It would be great if someone could help with any of these.
Thanks

If the picture doesnt appear, the questions are at: http://www.geocities.com/jan_jac/induction.jpg

 

[McLake]

Neither are working ...

 

[flyin']

All I have to say was 2002 HSC Extension 2 induction problem was the biggest joke ever~

 

[Harimau]

Originally posted by flyin'
All I have to say was 2002 HSC Extension 2 induction problem was the biggest joke ever~

What was it?

 

[McLake]

Originally posted by flyin'
All I have to say was 2002 HSC Extension 2 induction problem was the biggest joke ever~

Remind me what it was again ...

 

[flyin']

It was proving "de Moivre's theorem". It was so textbook. And to think I spent so much time, memorising and developing solutions for harder induction! *Grumbles*

 

[bobo123]

anyone got answers to those 3?

 

[spice girl]

I'm only gonna do the first one today

Well we know 1 + 2 + 3 + ... + n = n(n+1)/2

So RHS = 4 * {n(n+1)/2}^3 = (1/2) * n^3 * (n+1)^3

Now RTP: LHS = RHS for all n

n=1
LHS = 1 + 3 = 4
RHS = (1/2) * 1^3 * (1+1)^3 = 4
LHS = RHS

Assume result true n=k
i.e. LHS(k) = (1/2) * k^3 * (k+1)^3
LHS(k) + (k+1)^3 + 3(k+1)^5 = (1/2) * k^3 * (k+1)^3 + (k+1)^3 + 3(k+1)^5
= (1/2)* (k+1)^3 * (k^3 + 2 + 6(k+1)^2)
= (1/2)* (k+1)^3 * (k^3 + 6k^2 + 12k + 8)
= (1/2)* (K+1)^3 * (k+1+1)^3

thus true for n=k+1

 

[McLake]

I still can't read the question, the picture is a broken link ....

 

[ezzy85]

i dont know why the pic wont work. Ill try attaching it.

 

[McLake]

Originally posted by ezzy85
i dont know why the pic wont work. Ill try attaching it.

OK, the picture works now, answers come later ...

 

[spice girl]

Just a little thought for the 2nd question.

The number 8888....88 (n digits) = 8 * 1111....11 (n digits)
and the number 1111....11 (n digits) is a sum of a geometric progression 1 + 10 + 100 + 1000 + ... + 10^(n-1)

 

[KeypadSDM]

Nearly got number 2 out:

For n = 1
RHS = (8/81)/[10^2 - 9 - 10]
= (8/81)[81]
= 8
= LHS
True for n=1

Assume true for n = k
S_k = (8/81)/[10^(k+1) - 9k - 10]

For n = k + 1

S_k + T_(k+1) = (8/81)[10^(k+1) - 9k - 10) + 888...888{to k + 1 places}

But 888...888{to k + 1 places}
= 8(1 + 10 + 100 + 10^k)
= 8[10^(k+1) - 1]/[10-1]
= (8/81)[9 * 10^(k+1) - 9]

(Thanks for the help spice_girl)

.: S_k + T_(k+1) = (8/81)[10^(k+1) - 9k - 10) + (8/81)[9 * 10^(k+1) - 9]
= (8/81)[10^(k+1) - 9k - 10 + 9 * 10^(k+1) - 9]
= (8/81)[10^(k+1) + 9 * 10^(k+1) - 9k - 9 - 10]
= (8/81)[10^(k+2) - 9(k + 1) - 10]
= S_(k+1)
.: true for n = k+ 1 if true for n = k
.: true for all integral n>= 1

 

[KeypadSDM]

The third one is evil because i clearly can't see the 10 line solution for n = k+1...

it took me 7 lines to simplify for n = 1 for god's sake....

 

[bobo123]

it looks kinda similar to the complex number Q i posted a while ago i think
but i still cant do it

 

[Harimau]

The trig induction one is so hard... i got up to:

RHS= [sin(k+1)A.cos(A/2) + cos(k+1)A.sin(A/2) - sin (A/2)] 2sin(A/2)

Trying to prove RHS= cos(k+1)A

Damn im so hopeless....

 

[spice girl]

3rd Q:

rewrite RHS to = sin({2n+1}@/2) - sin(@/2) / 2sin(@/2)

from sums-to-products this = [2sin(n@/2)cos({n+1}@/2)] / 2sin(@/2)
= sin(n@/2)cos({n+1}@/2) / sin(@/2)

Case n=1
LHS = cos@
RHS = sin(n@/2)cos({n+1}@/2) / sin(@/2)
= sin(@/2)cos(2@/2)/sin(@/2) = cos@ = LHS

Suppose result true for n=k
i.e. LHS(k) = RHS(k)
=> cos@ + cos2@ + ... + cosk@ = sin(k@/2)cos({k+1}@/2) / sin(@/2)

then
{cos@ + cos2@ + ... + cosk@} + cos(k+1)@ = {sin(k@/2)cos({k+1}@/2) / sin(@/2)
} + cos(k+1)@

now RHS here
= [1/sin(@/2)]*{sin(k@/2)cos({k+1}@/2) + sin(@/2)cos({k+1}@)}

now note that: cos({k+1}@/2) = cos({k+2}@/2)cos(@/2) + sin({k+2}@/2)sin(@/2)
and also: cos({k+1}@) = cos(k@/2)cos({k+2}@/2) - sin(k@/2)sin({k+2}@/2)

so RHS now becomes:
= [1/sin(@/2)]*{sin(k@/2)cos({k+2}@/2)cos(@/2) + sin(k@/2)sin({k+2}@/2)sin(@/2) + sin(@/2)cos(k@/2)cos({k+2}@/2) - sin(@/2)sin(k@/2)sin({k+2}@/2)}
= [1/sin(@/2)]*{sin(k@/2)cos({k+2}@/2)cos(@/2) + sin(@/2)cos(k@/2)cos({k+2}@/2)}
= [1/sin(@/2)]*)]*{[sin(k@/2)cos(@/2) + sin(@/2)cos(k@/2)]cos({k+2}@/2)}
= [1/sin(@/2)]*)]*{sin({k+1}@/2)cos({k+2}@/2)}
= RHS(k+1)

Thus LHS(k+1) = RHS(k+1)

I'd like to know an easier way to do this!

Anyway, the sums-to-products formula I used was: using P = A + B, Q = A - B
sinP - sinQ = sin(A+B) - sin(A-B) = 2sinBcosA = 2sin({P-Q}/2)cos(P+Q}/2)

 

[OLDMAN]

There is a slightly easier way, using products-to-sums instead. Maintain the orig. RHS:
sin((2k+1)@/2)/2sin(@/2)-1/2

RHS + cos((k+1)@)=sin((2k+1)@/2)/2sin(@/2)-1/2+cos((k+1)@)
={sin((2k+1)@/2)+2sin(@/2)cos((k+1)@)}/2sin(@/2) -1/2=
{sin((2k+1)@/2)+sin((k+1+1/2)@)-sin((k+1/2)@)}/2sin(@/2) -1/2
=sin((2(k+1)+1)@/2)/2sin(@/2)-1/2=RHS(k+1)

How does one memorize all these sums-to-products/products-to-sums formulae. Perhaps the following might help as a memory aid.

Any two numbers A,B could always be written as a sum and difference of their midpoint(A+B)/2 and deviation(A-B)/2. eg. 4, 2 could be written as 3+1,3-1; 8,4 as 6+2,6-2.
Thus sin(4)+sin(2)=sin(3+1)+sin(3-1)=2sin(3)cos(1)
cos(8)-cos(4)=cos(6+2)-cos(6-2)=-2sin(6)sin(2)

 

[maniacguy]

The annoying thing is that there are easier ways than induction if you're just asked to prove these. For example:

No. 2:

LHS = SUM[i=1...n, 888...888 (i places) ]
= 8/9 * SUM[i=1...n, 999...999 (i places) ]
= 8/9 * SUM[i=1...n, (10^i - 1)]
= 8/9 * [SUM[i=1...n, 10^i] - n]
= 8/9 * [10*(10^n - 1)/9 - n]
= 8/81 * [10^(n+1) - 9n - 10]
= RHS

If you see a place to insert a simple induction in here, do so - the question says prove by induction. If it's not explicit, then as long as you include one induction that should be okay (in theory, anyway)

One method to do number three involves noting that the LHS is simply the sum of the real parts of cis(n*theta), so that it must be the real part of the sum of cis(n*theta) and going from there.