When a problem gives you an equation and tells you to find the first four terms...

[abel muroi]

Here is the equation...

a_n = 1/n+1 - 1/n+2 (both of these are fractions, ex: 1/n+1 is a fraction)


and here is the work that i did...

=1/1+1 - 1/1+2
=1/2 - 1/3
=1/6 (is this how you find the first term of the equation?)

 

[Steven G]

Here is the equation...

a_n = 1/n+1 - 1/n+2 (both of these are fractions, ex: 1/n+1 is a fraction)


and here is the work that i did...

=1/1+1 - 1/1+2
=1/2 - 1/3
=1/6 (is this how you find the first term of the equation?)

You clearly know how to use parenthesis as you used them. You also knew that you needed them for the denominator so why not use them? a_n = 1/(n+1) - 1/(n+2).

=1/1+1 - 1/1+2--> what does blank space = 1/1+1 - 1/1+2 mean?
say what you mean. a_1 = 1/(1+1) - 1/(1+2)=1/2 - 1/3 =1/6 which is correct. Good for you.

 

[stapel]

Here is the equation...

a_n = 1/n+1 - 1/n+2 (both of these are fractions, ex: 1/n+1 is a fraction)

I will guess that the underscores indicate subscripts, so "a_n" means a_n. As posted, the right-hand side means this:

. . . . .\(\displaystyle \dfrac{1}{n}\, +\, 1\, -\, \dfrac{1}{n}\, +\, 2\)

I will guess that you neglected to include the grouping symbols, meaning instead to have posted "1/(n + 1) - 1/(n + 2)", which typesets as:

. . . . .\(\displaystyle \dfrac{1}{n\, +\, 1}\, -\, \dfrac{1}{n\, +\, 2}\)

and here is the work that i did...

Where is the rest of the exercise? We first need the instructions. Without that information, we have no way of knowing if you've done "it" correctly, since we don't know what "it" is that you're supposed to be doing with the initial sequence-term formula.

=1/1+1 - 1/1+2

This starts with an "equals" sign. What is this equal to? What did you start with? What are you doing?

Please be complete. Thank you!

 

[HallsofIvy]

Here is the equation...

a_n = 1/n+1 - 1/n+2 (both of these are fractions, ex: 1/n+1 is a fraction)

PLEASE use parenthese! You mean a_n= 1/(n+1)- 1/(n+2) what you wrote, 1/n+ 1- 1/n+ 2, is equal to (1/n- 1/n)+ 1+ 2= 3 for all n.

and here is the work that i did...

=1/1+1 - 1/1+2
=1/2 - 1/3
=1/6 (is this how you find the first term of the equation?)

Yes, that is the first term. Can you do the same thing to find the nth term?

 

[abel muroi]

PLEASE use parenthese! You mean a_n= 1/(n+1)- 1/(n+2) what you wrote, 1/n+ 1- 1/n+ 2, is equal to (1/n- 1/n)+ 1+ 2= 3 for all n.


Yes, that is the first term. Can you do the same thing to find the nth term?

Isn't finding the nth term just trying to find a formula that will find ANY term in a sequence right?

So if that's true, then the nth term is just

[FONT=MathJax_Main]a_{n =} [/FONT]1/(n + 1) - 1/(n + 2)[FONT=MathJax_Main][/FONT]

 

[Steven G]

Isn't finding the nth term just trying to find a formula that will find ANY term in a sequence right?

So if that's true, then the nth term is just

[FONT=MathJax_Main]a_{n =} [/FONT]1/(n + 1) - 1/(n + 2)

I would do the subtraction and get a_n = 1/[(n+1)(n+2)]

Note how much easier it is to find a_1 using the new formula

 

[abel muroi]

I would do the subtraction and get a_n = 1/[(n+1)(n+2)]

Note how much easier it is to find a_1 using the new formula

I have a question

Using the formula a_n = 1/(n + 1) - 1/(n + 2) and your formula, I got the first four terms of the equation

a_{1 =} 1/6

a₂ = 1/12

a₃ = 1/20

a_{4 =} 1/30

My question is... how can i find the ratio/ difference here? I am not entirely sure if this sequence is an arithmetic or geometric

 

[Steven G]

I have a question

Using the formula a_n = 1/(n + 1) - 1/(n + 2) and your formula, I got the first four terms of the equation

a_{1 =} 1/6

a₂ = 1/12

a₃ = 1/20

a_{4 =} 1/30

My question is... how can i find the ratio/ difference here? I am not entirely sure if this sequence is an arithmetic or geometric

Not all sequences are either arithmetic or geometric. Consider 1,0,1,1,0,1,1,1,0,1,1,1,1,0,1,1,1,1,1,0..... or 1,-1,1,-1,1,-1,....


Is it arithmetic? compute a_n - a_(n-1) = 1/[(n+1)(n+2)] -1/[n(n+1)] then simplify it to see that it is not a constant (actually simplify). So not arithmetic.

Is it geometric? Compute a_n/a_(n-1) = {1/[(n+1)(n+2)]}/ {1/[n(n+1)] } = n/(n+2) which is not constant. So not geometric.