Sum of series (1 Viewer)

[QZP]

Hi, I encountered this question that I cannot do:

The nth term of a series is given by Tn = 1/(2n-1)(2n+1). Explain why the sum of the first n terms of the series is n/(2n+1).

So I listed Sn = 1/3 + 1/15 + 1/35 + ... + 1/(2n-1)(2n+1)

There is no common ratio/difference :S

Then I tried to split it up into two separate series of Tn1 = 1/(2n-1) and Tn2 = 1/(2n+1)
Sn1 = 1/1 + 1/3 + 1/5 + ... + 1/(2n-1)
Sn2 = 1/3 + 1/5 + ... + 1/(2n+1)

But these two separate series have no common difference/ratio such that I can do Sn1 times Sn2 to find Sn.

Help appreciated D:

 

[HeroicPandas]

where did you get this question from? (i need to know because i can see a method, but its not 3U)

 

[QZP]

where did you get this question from?

http://www.hsc.csu.edu.au/maths/ext1/math_induction/179/mathinduc.htm

Part 3 Q1.c

 

[Sy123]

Hi, I encountered this question that I cannot do:

The nth term of a series is given by Tn = 1/(2n-1)(2n+1). Explain why the sum of the first n terms of the series is n/(2n+1).

So I listened Sn = 1/3 + 1/15 + 1/35 + ... + 1/(2n-1)(2n+1)

There is no common ratio/difference :S

Then I tried to split it up into two separate series of Tn1 = 1/(2n-1) and Tn2 = 1/(2n+1)
Sn1 = 1/1 + 1/3 + 1/5 + ... + 1/(2n-1)
Sn2 = 1/3 + 1/5 + ... + 1/(2n+1)

But these two separate series have no common difference/ratio such that I can do Sn1 times Sn2 to find Sn.

Help appreciated D:

Now, try expanding that sum out, it doesn't have a common difference, but see what happens when you write that series out in the a+b+c+d+..... form

 

[HeroicPandas]

http://www.hsc.csu.edu.au/maths/ext1/math_induction/179/mathinduc.htm

Part 3 Q1.c

Are you trying to do part (c) without doing part (b)?

 

[QZP]

Are you trying to do part (c) without doing part (b)?

I've done it; i just decided a,b were irrelevant for c LOL :S

 

[HeroicPandas]

I had the same thought of doing that then it becomes a tele... but is can you use the method of partial fractions in 3U?

 

[Sy123]

I had the same thought of doing that then it becomes a tele... but is can you use the method of partial fractions in 3U?

Well its a very simple fraction split so technically its a 4U question but yea lol

 

[QZP]

Now, try expanding that sum out, it doesn't have a common difference, but see what happens when you write that series out in the a+b+c+d+..... form

Wowow! You get 1/2 (2n/(2n+1)) = n/(2n+1)

That is such an amazing result. How did you see it? Or is it something common... (looking at your 4u discussion). I'm somewhat inexperienced in math so...

 

[RealiseNothing]

It's under mathematical induction...

 

[Sy123]

Wowow! You get 1/2 (2n/(2n+1)) = n/(2n+1)

That is such an amazing result. How did you see it? Or is it something common... (looking at your 4u discussion). I'm somewhat inexperienced in math so...

Its a topic called partial fractions in 4U

It's under mathematical induction...

This lol they would never ask a straight out series in 3U, let alone give it without some guidance in 4U