Induction question (1 Viewer)

[cutemouse]

Hey guys,

I have no idea on where to start for this one. Any help would be appreciated.

Thanks

 

[bilal.a]

do u go 2 uws, stats or eco methods?

 

[cutemouse]

do u go 2 uws, stats or eco methods?

???

 

[gurmies]

First part



Next part is very simple, I don't think I will waste my time using latex.

The expression on the LHS of the inequality is essentially the same as the expression I have proved above, however each term is multiplied by A_{n} (with decreasing subscript) and so on, instead of B_{n}. Therefore, expression multiplied by A_{n} <= B_{n}(As A_r <= B_r), but since expression multiplied by B_{n} = sqrt{b1} + sqrt{b2} +...+sqrt{bn}, we get the required result.

 

[Timothy.Siu]

for n=1

LHS= 1/rootb1 x B1 = root B1 = RHS

edit: lol he delted his post =(

 

[gurmies]

B_{1} and b_{1} are essentially the same.

LHS = B_{1}/sqrt[b_{1}] = sqrt[b_{1}]

RHS = sqrt[b_{1}] = LHS

EDIT: Oh, lol.

 

[cutemouse]

Yeah I realised what I had to do. *Hits self on head*

I'll try part (ii) later. But now I need some sleep. Oh if anyone wants to have a go at part (ii) then that'd also be appreciated .

 

[00iCon]

Every time I see questions like this, I get scared that I'm going to need to know the sequences and series formulae. I never learnt them properly! But it's always so easy!

 

[gurmies]

Every time I see questions like this, I get scared that I'm going to need to know the sequences and series formulae. I never learnt them properly! But it's always so easy!

Conrad, always attempt the question - you'll be surprised.

 

[cutemouse]

Hmm takers for part (ii) and (iii)?

I think I kinda got (ii) but not too sure. I don't really wanna post my 'scraps' of working up here when I don't know that it's correct, and it's messy too

 

[lolokay]

I haven't done anything else from the question, but for part ii)

they give you A_r = < B_r
just use that in conjunction with the other 2 identities you have (part i) and the one they just gave you)