induction q. from fitz. (1 Viewer)

[dilos]

hey, i'm just stuck in the process of the whole thing...just kinda getting used to induction- could someone write the solution (with the process) for this q. pleassse... from fitzpatric, 23(c)q 9

1+2+3^2+3^3+...+3^(n-1)= (3^(n) - 1) / 2

thanks!!!

 

[CM_Tutor]

Theorem: 1 + 3 + 3² + ... + 3^n-1 = (3ⁿ - 1) / 2, for all integers n=>1

Proof: By induction on n.

A Put n = 1: LHS = 3⁰ = 1
RHS = (3¹ - 1) / 2 = (3 - 1) / 2 = 1 = LHS
So, the result is true for n = 1.

B Let k be a value of n for which the result is true.
That is, 1 + 3 + 3² + ... + 3^k-1 = (3^k - 1) / 2 ________(**)

(Comment: Note that the above statement is called the induction hypothesis. It is labelled (**) because it is very important)

We must now prove the result for n = k + 1.
That is, we must prove 1 + 3 + 3² + ... + 3^k = (3^k+1 - 1) / 2

LHS = 1 + 3 + 3² + ... + 3^k
= [1 + 3 + 3² + ... + 3^k-1] + 3^k
= [(3^k - 1) / 2] + 3^k, using the induction hypothesis (**)
= (1 / 2) * (3^k - 1 + 2 * 3^k)
= (3 * 3^k - 1) / 2
= (3^k+1 - 1) / 2
= RHS

So, if the result is true for n = k, then it is also true for n = k + 1.

C It follow from A and B by mathematical induction that the result is true for all integers n => 1.

 

[CM_Tutor]

Note that if you were asked to prove this result, but were not required to use induction to do so, then it would be much easier to simply sum the GP.

 

[Grey Council]

hrm, if my advice is worth anything, do alot of induction. alot and alot and alot. first of all, its interesting, and fun. but more importantly, it did wonders for my algebra. Induction and differentiation pretty much got my algebra not rusty. (lol, i won't say my algebra is good, cause its not, but its not rusty. )

 

[dilos]

thanks for your help guyz!!!

i'm so silly---stoopid algebraic mistakes throughout

 

[Grey Council]

lol, trust me, thats why i said do heaps of induction and differentiation. it'll sharpen your algebra.

 

[Xayma]

And all types of induction problems like unequalities etc.

 

[CM_Tutor]

Originally posted by Xayma
And all types of induction problems like unequalities etc.

And remember that factorial induction is a possibility, so make sure that you do some once you have covered the binomial theorem. I have a variety of induction problems, if anyone needs some more to practice.

 

[dilos]

arr....stuck on another one...

Sn= n/2 {2a+(n-1)d}


*confuzzled*

 

[~*HSC 4 life*~]

induction's soooooo fun!

what's the question dilos? are you meant to prove the formula for sum of n terms by math induction?

 

[CM_Tutor]

~*HSC 4 life*~, the question was no doubt something like:

Using mathematical induction, prove that

a + (a + d) + (a + 2d) + ... + [a + (n - 1)d] = (n / 2) * [2a + (n - 1)d]

where a and d are constant, for all positive integers n.

 

[dilos]

Originally posted by ~*HSC 4 life*~
induction's soooooo fun!

what's the question dilos? are you meant to prove the formula for sum of n terms by math induction?

yep!! can you help me out....i'm kinda stuck 3/4 of the way into the question (the factorising part)..

thanks dude

 

[dilos]

btw. this is another question but:

are

n(n+1) is an even number.

and

2+4+6+...+2n= n(n+1)

(proof by induction)

the same question? cos they're asked one after the other.

 

[CM_Tutor]

Originally posted by dilos
are

n(n+1) is an even number.

and

2+4+6+...+2n= n(n+1)

(proof by induction)

the same question? cos they're asked one after the other.

No, they aren't they are separate induction problems. ie.

1. Prove by mathematical induction that n(n + 1) is even (ie. a multiple of 2) for all positive integers n.

2. Prove by mathematical induction that 2 + 4 + 6 + ... + 2n = n(n + 1), for all positive integers n.

 

[~*HSC 4 life*~]

You would start the first one like:
prove
n(n+1)= 2P

prove true for n=1
LHS= 1(1+1)
= 2
which is divisble by 2

Assume true for n=k
etc etc

as for the Sn one, i got stuck too it's late and i'm tired and ive got ext maths 7:30 tommrow lol, i'll leave it to CM_Tutor coz he/she is a brain and a half

 

[CM_Tutor]

Theorem: a + (a + d) + (a + 2d) + ... + [a + (n - 1)d] = (n / 2) * [2a + (n - 1)d], for all integers n=>1

Proof: By induction on n.

A Put n = 1: LHS = a
RHS = (1 / 2) * [2a + (1 - 1)d] = 2a / 2 = a = LHS
So, the result is true for n = 1.

B Let k be a value of n for which the result is true.
That is, a + (a + d) + (a + 2d) + ... + [a + (k - 1)d] = (k / 2) * [2a + (k - 1)d] ________(**)
We must now prove the result for n = k + 1.
That is, we must prove a + (a + d) + (a + 2d) + ... + (a + kd) = [(k + 1) / 2] * (2a + kd)

LHS = a + (a + d) + (a + 2d) + ... + [a + kd]
= {a + (a + d) + (a + 2d) + ... + [a + (k - 1)d]} + (a + kd)
= {(k / 2) * [2a + (k - 1)d]} + (a + kd), using the induction hypothesis (**)
= 2ak / 2 + k(k-1)d / 2 + a + kd
= ak + a + (d / 2)* [k(k - 1) + 2k]
= (k + 1)a + d(k² - k + 2k) / 2
= 2(k + 1)a / 2 + dk(k + 1) / 2
= [(k + 1) / 2] * (2a + kd)
= RHS

So, if the result is true for n = k, then it is also true for n = k + 1.

C It follow from A and B by mathematical induction that the result is true for all integers n => 1.

 

[dilos]

thanks CM_tutor!! you rule....

errr...i'm ashamed to ask you another question...

how do you do this one:

n
(2r-1)^(3) = n^(2) [2n^(2)-1]
r=1

 

[CM_Tutor]

Theorem: 1³ + 3³ + 5³ + ... + (2n - 1)³ = n²(2n² - 1), for all integers n=>1

Proof: By induction on n.

A Put n = 1: LHS = 1³ = 1
RHS = 1²[2(1)² - 1] = 1 * (2 - 1) = 1 = LHS
So, the result is true for n = 1.

B Let k be a value of n for which the result is true.
That is, 1³ + 3³ + 5³ + ... + (2k - 1)³ = k²(2k² - 1) ________(**)
We must now prove the result for n = k + 1.
That is, we must prove 1³ + 3³ + 5³ + ... + (2k + 1)³ = (k + 1)²[2(k + 1)² - 1]

LHS = 1³ + 3³ + 5³ + ... + (2k + 1)³
= [1³ + 3³ + 5³ + ... + (2k - 1)³] + (2k + 1)³
= [k²(2k² - 1)] + (2k + 1)³, using the induction hypothesis (**)
= 2k⁴ - k² + (2k)³ + 3(2k)²(1) + 3(2k)(1)² + 1³
= 2k⁴ + 8k³ + 11k² + 6k + 1
RHS = (k + 1)²[2(k + 1)² - 1]
= (k² + 2k + 1)(2k² + 4k +1)
= 2k⁴ + 4k³ + 2k² + 4k³ + 8k² + 4k + k² + 2k + 1
= 2k⁴ + 8k³ + 11k² + 6k + 1
= LHS

So, if the result is true for n = k, then it is also true for n = k + 1.

C It follow from A and B by mathematical induction that the result is true for all integers n => 1.

-----

Comment: In B, it would be better to not have to expand both sides, but rather transform LHS into RHS, but I don't see an easy way to do that off hand.

 

[dilos]

hahaha, thanks dude


guess what....ur stuck with me and my stoopid q's!

what about...this one:

d/dx (x^n) = nx^(n-1) for all n=>1


do i suck or do i suck?
*anyone answers that- and they die*

are you annoyed at me yet CM_tutor...

 

[CM_Tutor]

This question was discussed on the other current induction thread - have a look, and then try it yourself.

And no, I'm not annoyed. Why do you ask?