Mathematical Induction (1 Viewer)

[mazza_728]

i know this was ages ago, i did this as my first topic in prelim Maths 3 unit but im doing some revision and its soo easy actually i think its the easiest topic of the whole course but im stuck on this one question:

1/(1.2)+1/(2.3)+1/(3.4)+.....+1/(n(n+1)) = 1 - 1/(n+1)

ps. the "." at short hand multiplication signs not decimal points.

Anyway go thru the steps and wats ur last few lines of working??

Ive got LHS= 1-1/(k+1) + 1/((k+1)(k+2)
=1 - (((k+2)+1)/(k+1)(k+2))

if only i could cancel the k+2's then i would have =RHS but i cant because of the plus sign... sorry i know its a bit confusing typing it up on here, but if u do know wat im talking about did u get the same or have i done something wrong somewhere??

tahnks xoxo

 

[CM_Tutor]

Originally posted by mazza_728
Ive got LHS= 1-1/(k+1) + 1/((k+1)(k+2)
=1 - (((k+2)+1)/(k+1)(k+2))

You've made a mistake with the minus in this step. It should be:

LHS = 1 - [1 / (k + 1)] + 1 / (k + 1)(k + 2)
= 1 - [1 / (k + 1)] * [ 1 - 1 / (k + 2)]
= 1 - [1 / (k + 1)] * [ (k + 2) - 1] / (k + 2)
= 1 - [1 / (k + 1)] * (k + 1) / (k + 2)
= 1 - 1 / (k + 2)
= RHS

 

[mazza_728]

Im lost!

why are minusing 1/(k+2) from 1??? isnt that just in the first half of the equation>>>or am i just stupid? and why did u remove the (k+1) denominator from the third term it was 1/(k+1)(k+2) and somehow it became 1-1/(k+2)???

My brain must of shut down.

 

[nike33]

no he has taken 1 / k+1 out as a common factor

1-1/(k+2) what is wrong with this? if your misinterpriting it, it may be easier to comprehend as 1-[1/(k+2)]

 

[mazza_728]

Is it 1-(1/k+1+1/(k+1)(k+2)??
so u have to change the 2nd term in the bracket into a minus too??
I understand the taking the 1/k+1 out as a common factor... i think i might get it just a lil confused about the signs?

Ps do either of you tutor becaue ur both very capable and smart!

 

[ND]

I'm not sure what 1-(1/k+1+1/(k+1)(k+2) is supposed to be (your bracketing was quite poor).

To see why it's negative consider this:

1-(1/a)+(1/ab)=1-(1/a)(1-(1/b))
if it were a + instead of a -, the RHS would expand out to be 1-(1/a)-(1/ab)

 

[mazza_728]

um sorry about bracketing 1-[1/(k+1) +1/(k+1)(k+2)] any betta??

I dont understand the lil a / b things??

 

[ND]

Your bracketing is still uneven (i.e. you've got more opens than closes). The a b thing is just to show you why it's negative (isn't that what this whole problem is about?). It's the same as the question, just let a=k+1, b=k+2.

 

[Heinz]

Maybe this will be easier to see.

 

[nike33]

haha i probably could tutor but im in yr12 ask CM!

 

[mazza_728]

tried to fix bracketing?? any help?? i understand heinz way but it seems a lil un hsc like ... u cheated well heinz can someone explain the other way in laymens terms?

 

[mazza_728]

shit nike33 u mean im competing wit u .. im screwed

 

[nike33]

heinz way in not nessacary as you have the 1- out the front to start. if you dont understand CMs method heres an outline

LHS = 1 - [1 / (k + 1)] + 1 / (k + 1)(k + 2)
= 1 - [1 / (k + 1)] * [ 1 - 1 / (k + 2)] --take 1/(k+1) out as a common factor
= 1 - [1 / (k + 1)] * [ (k + 2) - 1] / (k + 2) --convert [ 1 - 1 / (k + 2)] to (k + 1) / (k + 2)
= 1 - [1 / (k + 1)] * (k + 1) / (k + 2) ..the k+1 cancel and you get the required answer! do all the cambridge 4u intergration qns (i think u said u did 4u somewhere and you will be set!)
= 1 - 1 / (k + 2)
= RHS

 

[mazza_728]

= 1 - [1 / (k + 1)] * [ 1 - 1 / (k + 2)] --take 1/(k+1) out as a common factor

why isnt this 1-[1/(k+1)]*[1+1/(k+2)]????

 

[ND]

Cos if you expand that you get 1-1/(k+1)-1/(k+1)(k+2).

 

[mazza_728]

i think i see it ... i think god i feel stupid

thanks guys..