Question 65997: here's the sequence/pattern:
1, -1, 2, -2, 3,....
how would u explain the pattern besides the fact that it goes in order and after the positive number you put the opposite of it?? or is that all?? or is some addition/subtraction involved??
Answer by Edwin McCravy(20056) (Show Source):
You can put this solution on YOUR website! here's the sequence/pattern:
1, -1, 2, -2, 3,....
how would u explain the pattern besides the fact that
it goes in order and after the positive number you put
the opposite of it?? or is that all?? or is some
addition/subtraction involved??
Let's find the general term for an
a1 = 1, a3 = 2, a5 = 3, and if n is odd, an = (n+1)/2
a2 = -1, a4 = -2, a6 = -3, and if n even, an = -n/2
So when n is odd, we want 1 times the formula (n+1)/2
plus 0 times the formula -n/2,
and when n is even, we want 0 times the formula (n+1)/2
plus 1 times the formula -n/2.
To do that, we make use of 2 special sequences
that alternate 0's and 1's:
1, 0, 1, 0, 1, 0, 1, ...
which has general term [1 + (-1)n+1]/2
and this sequence
0, 1, 0, 1, 0, 1, 0, ...
which has general term [1 + (-1)n]/2
We showed above that
if n is odd, an = (n+1)/2
and if n is even, an = -n/2
So when n is odd, we want 1 times
the "odd" formula plus 0 times the
"even" formula,
and
when n is even, we want 1 times
the "even" formula plus 0 times the
odd "formula".
We can do that by multiplying
the general term of the 1,0,1,0,...
sequence by the odd formula
and the general term of the
0,1,0,1,... sequence by the even
formula, and adding them
1+(-1)n+1 n+1 1 + (-1)n -n
-----------·----- + -----------· ---
2 2 2 2
Multiplying numerators and denominators,
use FOIL on first and distribute on
second
n + 1 + (-1)n+1n + (-1)n+1 -n - (-1)nn
--------------------------- + -------------
4 4
Combine all over the LCD of 4
n + 1 + (-1)n+1n + (-1)n+1 - n - (-1)nn
------------------------------------------
4
The n and -n cancel
1 + (-1)n+1n + (-1)n+1 - (-1)nn
---------------------------------
4
The last term of the numerator -(-1)nn can
be written as +(-1)n+1n
1 + (-1)n+1n + (-1)n+1 + (-1)n+1n
------------------------------------
4
1 + 2(-1)n+1n + (-1)n+1
-------------------------
4
Factoring (-1)n+1 out of the last
two terms.
1 + (-1)n+1(2n + 1)
an = ---------------------
4
That's your formula for an, the nth term
Edwin
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