HGH's '01 trial q.10c (1 Viewer)

[Hatta]

This question's been bugging me for ages. The 2001 Hornsby Girl's trial ( http://www.boredofstudies.org/courses/maths/2u/2001_Maths2_T_Hornsby_q.pdf ), question 10 c) is:

Observe that:
1= 1
3x= x + 2x
5x^2 = x^2 + 2x^2 + 2x^2
7x^3 = x^3 + 2x^3 + 2x^3 + 2x^3
9x^4 = x^4 + 2x^4 + 2x^4 + 2x^4 + 2x^4

By studying the above arrangement, or otherwise, find in simplest algebraic form, an expression for the limiting sum of the series:
1 + 3x + 5x^2 + 7x^3 + 9x^4 + .... + (2n-1)x^(n-1) +...

I've attacked this question so much I'm ready to scream. I looked at the answers, and couldn't figure out how they got what they got.

Could someone please help explain it to me?

 

[lyounamu]

This question's been bugging me for ages. The 2001 Hornsby Girl's trial ( http://www.boredofstudies.org/courses/maths/2u/2001_Maths2_T_Hornsby_q.pdf ), question 10 c) is:

Observe that:
1= 1
3x= x + 2x
5x^2 = x^2 + 2x^2 + 2x^2
7x^3 = x^3 + 2x^3 + 2x^3 + 2x^3
9x^4 = x^4 + 2x^4 + 2x^4 + 2x^4 + 2x^4

By studying the above arrangement, or otherwise, find in simplest algebraic form, an expression for the limiting sum of the series:
1 + 3x + 5x^2 + 7x^3 + 9x^4 + .... + (2n-1)x^(n-1) +...

I've attacked this question so much I'm ready to scream. I looked at the answers, and couldn't figure out how they got what they got.

Could someone please help explain it to me?

If you looked at my beautifully coloured parts, they are actually in series (infinitely).

And that patterns continue throughout.

So if you make a separte infinit series sum:

S = 1/(1-x) + 2x/(1-x) + 2x^2/(1-x) + .... infinitely
= 1/(1-x) (1 + 2x + 2x^2 + ....)
= 1/(1-x) (1 + (2x + 2x^2 +...))
= 1/(1-x) ( 1+ 2x/(1-x))
= 1/(1-x) ( 1+x)/(1-x)
= (1+x)/(1-x)^2 #

 

[Hatta]

Ah, that makes it clearer. Thank you!

 

[Wolfbeard]

Thanks. That helps me, too.