Question 115576: I am very frustrated! I am doing a piece of coursework and have discovered a changing difference pattern. Could someone please explain to me how to apply it and simplify it for my own results. I know the algebraic expression for a changing difference sequence is: a + (n-1)d + half(n-1)(n-2)c
a = value of first term in sequence
d = first difference between the first two numbers
c = is the change between one difference and the next
a = 481
d = 600
c = 240
Changing Difference Sequence:
481, 1081, 1921, 3001, 4321 Sequence
600 840 1080 1320 first difference
240 240 240 Second difference
I have tried for several hours to figure it out for my own results but have just got frustrated. I need to solve the nth term formula for this changing difference sequence to move on with my coursework. Thank you so much for any help!
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a + (n-1)d + half(n-1)(n-2)c
a = value of first term in sequence
d = first difference between the first two numbers
c = is the change between one difference and the next
a = 481
d = 600
c = 240
Changing Difference Sequence:
481, 1081, 1921, 3001, 4321 Sequence
600 840 1080 1320 first difference
240 240 240 Second difference
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changing difference = 481+(5-1)600 + (1/2)(5-1)(5-2)240
cd = 481 + 2400 + 6*240
cd = 4321
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Comment: I've never heard of this "changing difference", but if your
formulas and numbers are correct 4321 is the result.
Question: What does cd=4321 tell you?
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Cheers,
Stan H.
Answer by richwmiller(17219)  (Show Source):
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