(adsbygoogle = window.adsbygoogle || []).push({}); 2 Induction Questions

1. The problem statement, all variables and given/known data

I have quite a few questions and so i just made it an image. Also attached.

http://img411.imageshack.us/img411/1002/inductionforlife.jpg" [Broken]

Only need help with questions2and5now

Oh and so far my lecturer has taught well-ordering, strong induction and simple induction. But I could only follow simple induction... the other two I'm quite clueless about >< Though tell me which method is best for each question.

2. Relevant equations

Refer to image

3. The attempt at a solutions

Question 2

I have no clue to how to start it..

All i have done is

24 = 7 + 7 + 5 + 5

25 = 5 + 5 + 5 + 5 + 5

26 = 7 + 7 + 7 + 5

27 = 7 + 5 + 5 + 5 + 5

28 = 7 + 7 + 7 + 7

29 = 7 + 7 + 5 + 5 + 5

no idea what to do now

[STRIKE]Question 3

I'm not sure if my method is correct but I've proved that when x = 0 and n = 1, x = 1 n = 2 are true. But I get stuck whilst proving n = k + 1

Let S_{n}be (1 + x)^{n}>= 1 + nx

For n = 1 and x = 0, S_{1}=

LHS = (1+0)^{1}= 1

RHS = 1 + (1)(0) = 1

Therfore LHS >= RHS Hence n = 1 is true.

Assume n = k is true

S_{k}--> (1 + x)^{k}>= 1 + kx

For n = k + 1, S_{k+1}=

I know that I need to get to

(1 + x)^{k+1}>= 1 + (k+1)x

(1 + x)^{k}>= 1 + kx

(1 + x)^{k}(1 + x)^{1}>= (1 + kx)(1 + x)^{1}(multiplied both sides by (x + 1)

(1 + x)^{k+1}>= 1 + x + kx + kx^{2}

I can see that on the RHS there is 1 + x + kx I'm not sure what to do with it... hints/help?

(Just needs checking)Question 4

Let S_{n}be ƒ_{1}+ ƒ_{2}+ ... + ƒ_{n}= ƒ_{n+2}-1

For n = 1, S_{1}

LHS = ƒ_{1}= 1

RHS = ƒ_{1+2}- 1 = 2 - 1 = 1

LHS = RHS

Therefore n = 1 is true

Assume true for n = k

S_{k}--> ƒ_{1}+ ƒ_{2}+ ... + ƒ_{k}= ƒ_{k+2}-1

For n = k + 1, S_{k+1}=

RHS = ƒ_{k+3}- 1

LHS = ƒ_{1}+ ƒ_{2}+ ... + ƒ_{k}+ ƒ_{k+1}

= ƒ_{k+2}-1 + ƒ_{k+1}

= ƒ_{k+2}+ ƒ_{k+1}- 1

= ƒ_{k+3}- 1 (should I write any reason here? if yes..what should i write?)

= RHS

Hence n = k + 1 is true

By mathematical induction S_{n}is true for all positive integers n.

[/STRIKE]

Question 5

Show that n/t - 1/(q+1) is positive and numerator is less than n

where t = nq + r with 0 < r < n

(get common denominator then expand and simplify)

n/t - 1/(q+1)

= n(q + 1)/[t(q+1)] - t/[t(q+1)]

= [n(q+1) - t] / [t(q+1)]

= [nq - t + n] / [t(q+1)]

t = nq + r

nq - t = -r

hence n/t - 1/(q+1)

= [n-r] / [t(q+1)]

from 0 < r < n

n > r therefore n - r > 0 (proved that numerator is positive)

and since r > 0 then n - r < n (proved that numerator is < n)

I'm not sure where to go from here

Please someone help me however you can..

Thank you in advance!!

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# Homework Help: Several Induction questions

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