Series (1 Viewer)

[hscwav2012]

This series question:
Find limiting sum???

1/5 + 2/5^2 + 3/5^3 + 4/5^4

 

[RealiseNothing]

Break it up into:



Now sum each series individually.

The first one is:



Second one:



Third one:



If we keep doing this, you notice that we form another GP?

So the sum of this GP is:

 

[bleakarcher]

I saw that Rivalry..

 

[Alkenes]

Nice solution mate

break it up into:



now sum each series individually.

The first one is:



second one:



third one:



if we keep doing this, you notice that we form another gp?

So the sum of this gp is:

 

[RivalryofTroll]

I saw that Rivalry..

lol i just realised i read the question wrong LOL

i was like why was this in the MX2 section xD

then re-read the question

 

[juantheron]

 

[Trebla]

Another approach:

If -1 < x < 1 then

 

[fortune_cookie]

Another approach:

If -1 < x < 1 then

not bad Oo

 

[bleakarcher]

Another way could be this:

(1/5)+(2/5^2)+(3/5^3)+...
=[(1/5)+(1/5^2)+(1/5^3)+...]+(1/5)[(1/5)+(1/5^2)+(1/5^3)+...]+(1/5^2)[(1/5)+(1/5^2)+(1/5^3)+...]+...
=[(1/5)+(1/5^2)+(1/5^3)+...][1+(1/5)+(1/5^2)+(1/5^3)+...]
=[(1/5)/[1-(1/5)]][1+(1/4)]
=(5/4)(1/4)=5/16

Damn, I wish I knew latex..

 

[fortune_cookie]

Another way could be this:

(1/5)+(2/5^2)+(3/5^3)+...
=[(1/5)+(1/5^2)+(1/5^3)+...]+(1/5)[(1/5)+(1/5^2)+(1/5^3)+...]+(1/5^2)[(1/5)+(1/5^2)+(1/5^3)+...]+...
=[(1/5)+(1/5^2)+(1/5^3)+...][1+(1/5)+(1/5^2)+(1/5^3)+...]
=[(1/5)/[1-(1/5)]][1+(1/4)]
=(5/4)(1/4)=5/16

Damn, I wish I knew latex..

That's basically what RealiseNothing did.

I think the method of writting out the limiting sum formula, differentiating and multiplying by x is by far the best method.

Also, if they were to change the question to 1/5 - 2/5^2 + 3/5^3 -4/5^4 +... (to have alternating + & - signs)

You can easily copy the method and just start with

1-x+x^2-x^3 +x^4 - ..... = 1/(1+x)

 

[iSplicer]

Another approach:

If -1 < x < 1 then

me gusta.