What is the equation for the sequence 1, 2, 1/4, 8, 1/16, 32,......
I see
2^0 , 2^1 , 2^-2 , 2^3 , 2^-4, 2^5 , ...
Try to state a general term for this.
let me know what you got.
how do i write that in terms of An? like An= (2)^n(-1)^n
some thing like that? or is that right
first of all, notice your primary exponent of n has to be one less than the term number
so lets start with 2^(n-1)
to give us 1,2,4,8,....
now you want to make all odd-numbered terms to have a negative exponent
Your (-1)^n does that perfectly.
so you were very close,
Term(n) = 2^((n-1)(-1)^n)
To find the equation for a given sequence, it is important to observe the pattern between the terms. Let's analyze the sequence you provided: 1, 2, 1/4, 8, 1/16, 32, ...
At first glance, it may seem that there is no obvious pattern between the terms. However, if we look closer, we can notice a pattern related to the powers of 2.
The first term is 1, which can be written as 2^0 (since any number to the power of 0 is 1).
The second term is 2, which can be written as 2^1.
The third term is 1/4, which is equivalent to 2^-2 (since 2 raised to the power of -2 is equal to 1/2^2, which is 1/4).
The fourth term is 8, which can be written as 2^3.
The fifth term is 1/16, equivalent to 2^-4.
The sixth term is 32, which is equal to 2^5.
From these observations, we can conclude that the pattern involves raising 2 to a power that increases by 1 each time.
Therefore, the equation for the given sequence is 2^(n-1), where n represents the position of the term in the sequence.
For example, the third term in the sequence would be found by substituting n=3 in the equation: 2^(3-1) = 2^2 = 1/4.
Hence, the equation for the sequence 1, 2, 1/4, 8, 1/16, 32, ... is 2^(n-1).