Question 31731: The problem is to find the general equation for the sequence, {-1,3,-5,7,-9,11,-13,15...}. I know that -1+4=3;-1-4=-5; -1+8=7;-1-8=-9; etc...so I am guessing it involves 4*n since the difference between the numbers is 4,8,12,16,20,24,28 respectively. Also, the alternating signs indicate -1^n. I also wondered if it involved absolute value. I just haven't been able to put it together with the formula for an arithmetic sequence.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! The problem is to find the general equation for the sequence, {-1,3,-5,7,-9,11,-13,15...}. I know that -1+4=3;-1-4=-5; -1+8=7;-1-8=-9; etc...so I am guessing it involves 4*n since the difference between the numbers is 4,8,12,16,20,24,28 respectively. Also, the alternating signs indicate -1^n. I also wondered if it involved absolute value. I just haven't been able to put it together with the formula for an arithmetic sequence.
YOU ARE ON RIGHT TRACK...THIS IS AN ALTERNATING SEQUENCE WITH 2 FORMULAE FOR N TH.
TERMS
ODD TERMS ARE .....-1,-5,-9,-13....ETC
SO A=-1,CD=-5+1=-4.....SO TM FOR THIS IS TM=-1-4(M-1)=-1-4M+4=3-4M....WE USED M HERE INSTEAD OF CONVENTIONAL N WHICH IS RESEVED FOR THE ORIGINAL SERIES
SO WHEN N IS ODD OR FOR T(2N-1)..THIS IS THE FORMULA
T(2N-1)=3-4N
NOW EVEN TERMS ARE...3,7,11,15.......ETC....BUT THESE ARE STARTING FROM SECOND TERM IN THE SERIES
SO A=3,CD=4...........TR=3+4(R-1)=4R-1
SO WHEN N IS EVEN WE HAVE ...FOR T2N...THIS IS THE FORMULA
T2N=4N-1..................
THUS THE SERIES IS DEFINED BY 2 FORMULAE(LIKE WE DO FOR FUNCTIONS IN DIFFERENT RANGES)
T(2N-1)=3-4N AND T2N=4N-1.
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