Series and Sequences Question (2 Viewers)

[GaDaMIt]

ive proved 1/n + 2/n + 3/n ..... + n/n = (n+1)/2

and then it goes

hence find the sum of the first 300 terms of 1/1 + 1/2 + 2/2 + 1/3 + 2/3 + 3/3 + 1/4 + 2/4+ 3/4+ 4/4 .. etc

help please?

EDIT: Sorry typed up wrong denominator at the start =\

 

[vafa]

sub 300 in (n+1)/n

 

[GaDaMIt]

answers have 162

 

[GaDaMIt]

i meant first 300 terms...


and also another question, not maths related

why is BoS lagging so badly? is it just me or what?

 

[Riviet]

why is BoS lagging so badly? is it just me or what?

Yeah, the server has been lagging alot in the last few days; mainly due to sudden influx of '06ers online on the forums and posting in them.

 

[GaDaMIt]

NVM sorry if anyone answered - BoS lagging badly anyway .. stupid question .. anyway another Q..

Tn = 2n + 3 + 2^n is a mix of AP and GP. USe combination of AP and GP formulae to find Sn?

 

[Riviet]

The GP part is 2ⁿ and the AP part is 2n+3. Just apply sum of a GP and AP to each respectively to find S_n.

 

[GaDaMIt]

i just made a start on mathematical induction..

prove by mathematical induction for all positive integers n

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



is there a mistake with this question? to my understanding of induction so far i have to prove its true for n = 1...

LHS = 1, RHS = 2^(1+1) - 1 = 2^2 - 1 = 4 - 1 = 3.. so wats going on????

 

[Riviet]

For n=1, LHS = 2⁰ + 2¹
=1 + 2
=3
=RHS

 

[Riviet]

Assume true for n=k (k is a positve integer): 1 + 2 + 2² + ... + 2^k = 2^k+1 - 1
Required to prove true for n=k+1: 1 + 2 + 2² + ... + 2^k + 2^k+1 = 2^k+2 - 1

LHS=1 + 2 + 2² + ... + 2^k + 2^k+1
=2^k+1 - 1 + 2^k+1, by assumption
=2.2^k+1 - 1
=2^k+2 - 1
=RHS

.'. true for n=k+1.

[insert your conclusion here]

 

[Raginsheep]

Actually the value of n changes depending on the question. Gernerally it'll be for 0 or 1 but depending on restrictions, it might be 2, 3, 4...

In this case, n = 0 since the first term is 1 or 2⁰.

 

[GaDaMIt]

Prove by induction that the sum of the angles of a polygon with n sides is n - 2 straight angles
[HINT: Dissect the (k+1)-gon into a k-gon and a triangle]


??
i am n00b

 

[SoulSearcher]

This post should give you a start:

http://community.boredofstudies.org/571688/post-25.html

 

[GaDaMIt]

prove inequalities by induction .. 3^n > n^2 for n >= 2 (and also for n = 0 and 1)

wats up with that?

why not just have it as n >= 0?

 

[Riviet]

BIG EDIT:

If n=0, 3⁰=1, 0²=0 => true for n=0
If n=1, 3¹=3, 1²=1 => true for n=1
If n=2, 3²=9, 2²=4 => true for n=2

.'. It is true for n=0, 1, and 2

Assuming 3^k > k², k is a positive integer AND k>2

Required to prove 3^k+1 > (k+1)² ie 3^k+1-(k+1)² > 0

3^k+1-(k+1)²=3k²-(k+1)² by assumption

>3(2)²-(2+1)² since k>2 because we already proved it true for n=0 and 1

=12-9

=3

>0 as required

[insert conclusion here]

 

[GaDaMIt]

It works for n=0, 1 and 2 so I'm not sure why the question has that.

Assuming 3^k > k², k is a positive integer
Required to prove 3^k+1 > k²

3^k+1=3.3^k
>3.k² by assumption
>k² since k is a positive integer, as required

[insert conclusion here]

Should be n = k+1 on the RHS... quoting you with all the "sups" is too confusing ill just hope you understand what im saying.. you need to prove its greater than (k+1)^2 for the last bit

 

[Riviet]

Yeah, sorry about that, no wonder it seemed so easy, I'll try it again.

 

[GaDaMIt]

Also, wat point do u suggest to be tested as step 1? n = 0 or n = 2?

 

[Riviet]

Also, wat point do u suggest to be tested as step 1? n = 0 or n = 2?

In an exam, I'd just prove all three to be safe if the question wasn't clear with the initial value. It does happen to work for n=0 and upwards.

 

[SoulSearcher]

Well here's this part of the proof, although I don't know how correct that is.

Continuing on, need to show that 3^k+1 > (k+1)²
i.e. 3^k+1 - (k+1)² > 0
LHS = 3^k+1 - (k+1)²
= 3*3^k - (k²+2k+1)
> 3(k²) - (k²+2k+1), using the induction hypothesis
= 3k² - k² - 2k - 1
= 2k² - 2k - 1
= 1/2 * (k²-k-1/2)
= 1/2 * [(k²-k+1/4) -1/2 -1/4]
= 1/2 * [(k - 1/2)² -3/4]
= (k - 1/2)²/2 - 3/8
> 0, since k > 2

[insert conclusion]