Need help with a question (1 Viewer)

[James1489]

This is question 7(a)(i) in the 2003 HSC paper.

It's pretty hard to type so here is the link to the 2003 paper http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2003exams/pdf_doc/mathematics_03.pdf

Find the limiting sum of the geometric series
2 + 2/( \sqrt{2} +1) + 2/( \sqrt{2} +1)^2

I've subbed in the values into the limiting sum formula, so:
S(infinity) = 2/(1 - (1/( \sqrt{2} +1))

I can't figure out how to simplify that down.
Any help would be appreciated.

Thanks.

 

[Mattamz]

S(infinity) = 2/ [1 - 1/(1+√2)]
= [2 + 2√2]/[1+√2 - 1]
= (2 + 2√2)/√2
=√2 +2

 

[James1489]

Yeah thanks heaps but what I don't understand is how you got the denominator in the line
= [2 + 2√2]/[1+√2 - 1]

Like I knew I had to flip it and multiply instead of divide, so I can see how you got [2 + 2√2] on the top, but how'd you get [1+√2 - 1] on the bottom?

Cheers for the help

 

[watatank]

Yeah thanks heaps but what I don't understand is how you got the denominator in the line
= [2 + 2√2]/[1+√2 - 1]

Like I knew I had to flip it and multiply instead of divide, so I can see how you got [2 + 2√2] on the top, but how'd you get [1+√2 - 1] on the bottom?

Cheers for the help

i suggest you do not flip and multiply. what i think Mattamz did was multiply the whole fraction by (1+√2)/(1+√2)

so 2/ [1 - 1/(1+√2)] becomes

2(1+√2) / [(1+√2) - 1]

and so on.

 

[darkliight]

Maybe a little easier to follow ...

2/[1 - 1/(1+√2)]
= 2/[√2/(1+√2)] --- (Just simplifying the denominator here, ie, writing 1 - 1/(1+√2) over a common denominator and then (1 + √2 - 1) = √2)
= (2 + 2√2)/√2 --- (Inverting and multiplying)
=√2 +2 --- (Simplifying again)