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Topic review (newest first)
- mathsyperson
- 2005-12-01 08:56:26
The second one is trickier.
Let's churn out the first few values and see how they relate to each other.
C_0 = 1
C_1 = 3
C_2 = 1
C_3 = 3
and so on.
These give differences of 2, -2, 2, -2... so we need a function of n that gives that. That can be done by using a variation of the original formula.
C_(n+1) = C_n - 2*(-1)^C_n
- MathsIsFun
- 2005-12-01 07:23:09
A bit hard to write that "n" in the right spot in C-sub-n ... Perhaps if you write C_n.
Anyway, I can do the first one easily:
C_n = n²
(a) C_1 = 1
(b) C_n = C_(n-1) +2(n-1) + 1
Let's check:
C_1 = 1
C_2 = 1 + 2(2-2) + 1 = 4
C_3 = 4 + 2(3-1) + 1 = 9
etc ...
- coolwind
- 2005-11-30 23:33:44
Give a recursive definition
ex:
Cn=7n
(a)C1=7;
(b)Cn+1=Cn+7;
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problem:
1.Cn=n^2 2.Cn=2-(-1)^n ????
thanks!!