easy MI question (1 Viewer)

[pLuvia]

Prove bu MI where n is a positive integer

n² - 3n + 2 > 0 for n > 1

 

[Estel]

Regarding thread title: easy? Looks like you don't need help then....

(n+1)^2 - 3(n+1) + 2 = n^2 - 3n + 2 + (2n - 2) >= (2n - 2) >= 0

 

[pLuvia]

Err. Can you do the full setting out, I get confused with the inequality type questions

Are we suppose to prove that

The statement is true for all n>1??

 

[100percent]

let >= be the greater than and equal sign
n² - 3n + 2 >= 0 for n >= 1
prove n=1
1-3+2
=0
is >= 0
therefore it is true for n=1

assume n=k
k²-3k+2 >=0

prove n=K+1
=(k+1)²-3(k+1)+2
=k²+2k+1-3k-3+2
=(k²-3k+2)+(2k-3)
>=0+2k-3 (from assumption)
>0 since 2k-3 greater than 0

edit-stuffed up rearranging

 

[pLuvia]

Why did you add 2k+2?

That's the part I don't understand?

Can't you just simplify from here?

k²+2k+1-3k+1+2

 

[100percent]

Why did you add 2k+2?

That's the part I don't understand?

Can't you just simplify from here?

i didn't add, i rearranged, you don't simpify coz you need to use the assumption. and our assumption was k²-3k+2 >=0, so replacing k²-3k+2 by 0 makes it >=0+2k+2 it's the thing that turns the = sign into >= which is what we need

 

[pLuvia]

Isn't it (k²-3k+2)+(2k-2)?

 

[100percent]

Isn't it (k²-3k+2)+(2k-2)?

yeh, i stuffed up my algebra, let me fix

 

[pLuvia]

And another question

3³ⁿ + 2ⁿ⁺² is divisible by 5

 

[100percent]

And another question

3³ⁿ + 2ⁿ⁺² is divisible by 5

3^3n + 2^(n+2)
n=1
3^3+2^(1+2)
3^3+2^3
35
5*7
is divisible by 5

n=k
3^3k + 2^(k+2) = 5m where m is integer

n=k+1
3^3(k+1) + 2^(k+1+2)
3^(3k+3) + 2*2^(k+2)
3^(3k+3) +2(5m-3^3k) (from assumption and rearranging)
3^(3k+3) + 10m -2*3^3k
3³*3^3k-2*3^3k+10m
[3^3k](27-2)+10m
25*3^3k+10m
5(5*3^3k+2m)
5n where n is integer, this is true since k & m are both integers

 

[Jago]

Assume true for k

3^3k + 2^k+2 = 5M

2^k+2 = 5M - 3^3k ------------------ 1

Prove true for k+1

= 3^3k x 3³ + 2^k+2 x 2

= 3^3k x 27 + (5M - 3^3k) x 2 --------------------- from 1

= 5 x 2M + 3^3k x 27 - 3^3k x 2

= 5 [ 2M + 5 x 3^3k ]

Edit: bah beaten by almost 10 minutes, oh well, mine looks nicer

 

[100percent]

I'm just wondering if my proof would work. It's slightly different.

3³ⁿ + 2ⁿ⁺² div by 5
let n=1
3³ + 2³=35 .:. true for n=1

Assume true for n=k
3^3k + 2^k+2 = 5M (where M is a positive integer value)

Consider n = k+1
3^3(k+1) + 2^k+2
=27.3^3k + 2.2^k+2
=27(3^3k + 2^k+2) - 25.2^k+2 (using assumption)
=27(5M) - 25.2^k+2
=135M - 25.2^k+2
=5(27M - 5.2^k+2)

yep thats fine

 

[pLuvia]

I'm just wondering if my proof would work. It's slightly different.

3³ⁿ + 2ⁿ⁺² div by 5
let n=1
3³ + 2³=35 .:. true for n=1

Assume true for n=k
3^3k + 2^k+2 = 5M (where M is a positive integer value)

Consider n = k+1
3^3(k+1) + 2^k+2
=27.3^3k + 2.2^k+2
=27(3^3k + 2^k+2) - 25.2^k+2 (using assumption)
=27(5M) - 25.2^k+2
=135M - 25.2^k+2
=5(27M - 5.2^k+2)

My teacher did this way and I had no clue how she got the 25

=27(3^3k + 2^k+2) - 25.2^k+2 (using assumption)

What did you exactly do here, I can see you used the assumption but where did you get the 25 from?

 

[100percent]

=27.3^3k + 2.2^k+2 -------------------------1
=27(3^3k + 2^k+2) - 25.2^k+2 ----------2

look at 2, if you expand you get
27.3</sup>3k</sup> + 27.2^k+2 - 25.2^k+2

factorise 2^k+2
27.3^3k + 2^k+2[27-25]
27.3^3k + 2.2^k+2
=equation 1

it's just a simple trick, nothing to it
example
x(1+x)ⁿ

can be written as
(1+x-1)(1+x)ⁿ
=(1+x)ⁿ⁺¹-(1+x)ⁿ

 

[.ben]

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

 

[acmilan]

For n=1, LHS = d/dx x = 1
RHS = 1.x⁰ = 1

Hence it is true for n = 1

Assume true for n = k

d/dx x^k = kx^k-1

Need to prove true for n = k+1

ie. prove d/dx x^k+1 = (k+1)x^k

LHS = d/dx x^k+1
= d/dx x^k.x
= x d/dx x^k + x^kd/dx x (using product rule)
= kx^k-1.x + x^k (using n = k assumption)
= kx^k + x^k
= (k+1)x^k
= RHS

 

[.ben]

o product rule! thanks acmilan

 

[.ben]

also:

30. (x+y)^n > x^n+y^n
how come it doesn't work for n=1?

 

[Riviet]

also:

30. (x+y)^n > x^n+y^n
how come it doesn't work for n=1?

Because we are saying that (x+y)ⁿ is strictly greater than xⁿ+yⁿ, meaning they can't be equal to each other.

 

[pLuvia]

Oh that question my teacher said that the restrictions were

n > 2
y > 0
x > 0