for the following exercises , write an explicit formula for each sequence,
4 , 7, 12 , 19 , 28....
please help ...
*NOT MY ANSWER* https://answers.yahoo.com/question/index?qid=20120121184030AAuQ9fU ----------------- An = A(n-1) + (2n-1) A1 = 4 A2 = 4 + 3 = 7 A3 = 7 + 5 = 12 A4 = 12 + 7 = 19 A5 = 19 + 9 = 28 A6 = 28 + 11 = 39 A7 = 39 + 13 = 52 ...
i dont get it
@directrix please help
I think it is neither a geometric nor arithmetic sequence. Therefore, I am looking for a pattern. From Alex's work, 3 is added to the first term to get the second. Then, 5 is added to the second term to get the third term. To get the explicit formula, the task is to find a relationship between the position number of a term and the value of the term. For position 1, a_1 = 4 For position 2, a_2 = 7 I then looked a the sequence in this way: 1 1^2 + 3 2 2^2 + 3 3 3^2 + 3 4 4^2 + 3 5 5^2 + 3 That format produced the sequence terms. Explicit term: a_n = n^2 + 3
is there a certain formula ?
If you want to know the 50th term in the sequence, a_50 = the square of the position number 50 added to 3. a_50 = 50^2 + 3
The explicit term is the formula for the nth term: Explicit term: a_n = n^2 + 3 I don't know a formula for finding the nth term of this particular sequence. I looked for a pattern.
There are nth term formulas for arithmetic and geometric sequences.
hmm.. well i got the first part but i have trouble determining the pattern
That requires patience and practice. The more of these you do, the better you get at it.
oh , how about this one 1 , 1 , 4/3 , 2 , 16/5 ....
I do not see that sequence as being geometric or arithmetic. I rewrote the sequence to try to find a pattern; 1/1 , 2/2 , 4/3, 8/4 , 16/5 ,
The denominators are 1, 2, 3, 4, 5 which are the same as the position numbers. 1, 2, 4, 8, 16 (the numerators) are powers of what number? The 1 may throw you off. If so, look at 2, 4, 8, 16. They are all powers of what number? @marcelie
2 raised to what power = 2 2 raised to what power = 4 2 raised to what power = 8 2 raised to what power = 16 @marcelie
1, 2, 4, 8, 16 2^0 , 2^1, 2^2 , 2^3 , 2^4 numerators The denominators are 1, 2, 3, 4, 5 which are the same as the position numbers.
Explicit term: a_n = 2^(n-1) / n |dw:1448875224608:dw|
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