a) note that you just keep subtracting 2, so
an = 1+2(n-1) = 2n-1
b) Hmmm. Differences are
8, 2, 3, 9, ...
.-6, 1, 6, ...
....7, 5 ...
Not a quadratic. The terms are
8, 8(1/4), 2(6/4), 3(12/4), ...
The numerators are 1,6,12. Again, not much good there.
The sequence is very short, so there are lots of rules that can work on just these four numbers. Note that if we subtract 1 from the last two terms, they are the same as the first two terms. So, we might say
a1 = 8
a2 = 2
an+2 = 1+a5-n
Not very satisfying. Maybe you can pierce the gloom.
(c) Note that the values are symmetric about an axis between the 0's. That indicates a quadratic, with an axis of symmetry at x = 5/2. So, we get
y = a(x - 5/2)^2 + k
Since y(2)= 0 and y(1) = -2,
a/4 + k = 0
9a/4 + k = -2
a = -1, k=1/4
y = -(x - 5/2)^2 + 1/4
= -(x-2)(x-3)
or,
an = -(n-2)(n-3)
For each of the following sequences, find an explicit and a recursive description of the pattern. Assume sequences start from n = 1.
a. 1, -1, -3, -5... b. 8,2,3,9...
c. -2, 0, 0, -2, -6...
1 answer