Math1004 (1 Viewer)

[adamsaclown]

if anyone thinks they're particularly good with it here, could you please explain the solution to Q5ii) in the 2002 Paper (about solving a recurrence relation)???


In particular, how does
1/[(1-2z)^3]

become:
infinity
SIGMA (m+2 CHOOSE 2) x (2z)^m
m = 0


that is, from the 6th last line to the 5th last line of the solutions??

Thanks

 

[SeDaTeD]

Um, try writing 1/(1-2z) as an infinite series, then differentiate it twice. Then divide both sides by some constant to get 1/[(1-2z)^3] and the RHS should be something like that.

 

[adamsaclown]

thanks

the general rule appears to be:

1/[(1-z)^n] = sum from k=0 to infinity of [(k+n-1 choose n-1)z^k]

 

[llamalope]

adamsaclown... you wouldn't happen to have dreadlocks would you? weird question I know...