Question (1 Viewer)

[acmilan]

This is a trial question from a school that i cant remember. Anyone have a clue how to do it.

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) + ...

 

[Affinity]

hmm let's do it the otehrwise way..

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

differentiating boths sides

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

doubling both sides

2 + 4x + 6x^2 + 8x^3 + ... = 2/(1-x)^2

subtract the original identity

(2 + 4x + 6x^2 + 8x^3 + ...) - (1 + x + x^2 + ...) = 2/(1-x)^2 - 1/(1-x)

etcetera

 

[acmilan]

How did you get 1+x+x^2 + ... = 1/(1-x)?

Anyone know how to do it using the pattern?

 

[Affinity]

that the limiting sum of a converging geometric series..

let (1 + x + x^2 + ... + x^n) = K
Kx = (x + x^2 + ... + x^(n+1) )
K - Kx = [1 - x^(n+1)]
K(1-x) = [1 - x^(n+1)]
K = [1 - x^(n+1)] / [1-x]

and if you take limit n to infinity, x^(n+1) tends to 0.
K = 1/(1-x)

by the pattern?

just add LHS up and RHS up.

LHS = what you want
if we let K = 1 + x + x^2 ...
RHS = K + 2(x)K + 2(x^2)K + ...
RHS = 2K + 2xK + 2(x^2)K + ...

= 2 [1 + x + x^2 + x^3 + ...]*K - K
= 2K^2 - K

etc.