Sequencesss

[Sara1111]

I need help !
Question: If the sequence 1, 1/4, 1/9, 1/16, .... then a14
a. 3/196
b.1/196
c.2/196
d.5/196

 

[firemath]

So have you been taught to calculate sequences?

 

[Sara1111]

So have you been taught to calculate sequences?

not really I was taught basic sequences only

 

[firemath]

You could calculate the answer, but just look at the sequence. What is in the numerator of every fraction in the sequence?

 

[Sara1111]

You could calculate the answer, but just look at the sequence. What is in the numerator of every fraction in the sequence?
[/QUOTE

You could calculate the answer, but just look at the sequence. What is in the numerator of every fraction in the sequence?

1

 

[firemath]

Correct. Which of the answers now seems correct as a final response?

 

[Sara1111]

Correct. Which of the answers now seems correct as a final response?

Option B

 

[Steven G]

You need to look for two things in sequences. One is what is not changing. The answer, seriously, is the 1 in the numerator and the division line. Now you need to decide what is changing. Clearly the denominator is changing. So far we know that the 14th term (possible before reducing) will be in the form of \(\displaystyle \dfrac{1}{}\).

Now we look at the denominator and see that they are 1 (1 = 1/1), 4, 9 and 16. To do this problem you need to recognize some pattern in these number.

Do you see a pattern? What is the pattern? Please respond back.

 

[pka]

I need help !
Question: If the sequence 1, 1/4, 1/9, 1/16, .... then a14
a. 3/196
b.1/196
c.2/196
d.5/196

The sequence is \(a_n=n^{-2}=\frac{1}{n^2}\)

 

[Steven G]

The sequence is \(a_n=n^{-2}=\frac{1}{n^2}\)

That depends on whether the 1st term is a₀ or a₁. I forget where I heard that from

 

[Sara1111]

You need to look for two things in sequences. One is what is not changing. The answer, seriously, is the 1 in the numerator and the division line. Now you need to decide what is changing. Clearly the denominator is changing. So far we know that the 14th term (possible before reducing) will be in the form of \(\displaystyle \dfrac{1}{}\).

Now we look at the denominator and see that they are 1 (1 = 1/1), 4, 9 and 16. To do this problem you need to recognize some pattern in these number.

Do you see a pattern? What is the pattern? Please respond back.

I think the pattern is multiplying numbers ( 1, 2, 3, 4, etc.) with the same number.
i'm not sure though

 

[Steven G]

I think the pattern is multiplying numbers ( 1, 2, 3, 4, etc.) with the same number.
i'm not sure though

pka gave you the answer.

1 = 1^2, 4 = 2^2, 9 = 3^2 and 16 = 4^2.

 

[Dr.Peterson]

I think the pattern is multiplying numbers ( 1, 2, 3, 4, etc.) with the same number.
i'm not sure though

Yes, the pattern involves squaring, which is the same as multiplying a number by itself.

It also involves putting that in the denominator (dividing 1 by the square of the index, which is 1 for the first term, 2 for the second, and so on).

So the 14th term is 1/14², which you can calculate, as you did for post #7.

 

[Sara1111]

Yes, the pattern involves squaring, which is the same as multiplying a number by itself.

It also involves putting that in the denominator (dividing 1 by the square of the index, which is 1 for the first term, 2 for the second, and so on).

So the 14th term is 1/14², which you can calculate, as you did for post #7.

Yes I got it
Thank you !