write a formula for the apparent nth term of the sequence ( just 2 more of these types) 1, -1/2, 1/3, -1/4, 1/5 (the other one) 4, 2, 4/3, 1, 4/5

Question

write a formula for the apparent nth term of the sequence ( just 2 more of these types)  1, -1/2, 1/3, -1/4, 1/5 (the other one) 4, 2, 4/3, 1, 4/5

callisto · Answer

@lakisha2013 What have you got for this question?

anonymous · Answer

i have been questions but they are hard i have not got an answer the last time i did this was about 2 years ago or more

callisto · Answer

I meant what have you done for this question... It doesn't matter you get the answer or not, but at least show us what you've tried. It's easier to help when we know what you don't understand.

anonymous · Answer

i have been trying to find the patterns but i cannot seem to find them

callisto · Answer

1 = [(-1)^(1-1)] (1/1)
-1/2 = [(-1)^(2-1)] (1/2)
1/3 = [(-1)^(3-1)] (1/2)
-1/4 = ...?
1/5 =...?
Can you work it out now?

anonymous · Answer

no because when i did the third problem i did not get 1/3

callisto · Answer

Sorry, it's 
1/3 = [(-1)^(3-1)] (1/3)

anonymous · Answer

so the next one would be [(-1)^(4-1)](1/4)

callisto · Answer

Yes.

callisto · Answer

Now, can you write it in general terms?

anonymous · Answer

it would be( -1^(n-1))(1/n)

callisto · Answer

Yes!

anonymous · Answer

i have started on the other one i got that the first one is either subtracting 2 or dividing 2

callisto · Answer

So, for the other one, 
4    = 4/1
2    = 4/2
4/3 = 4/3
1     = ... ?
4/5  = ...?

anonymous · Answer

it would be 4/n

callisto · Answer

Yes!
Sorry.. I couldn't see your post. For the second question, it can hardly involve subtraction because of the denominator... (it varies!)

anonymous · Answer

ok thank

callisto · Answer

Welcome!