Question

Find a formula for the general term an of the sequence (starting with a1):
{−3/16,4/25,−5/36,6/49,…}. I'm having problems dealing with the negative signs...

Answer 1

Few hints:

This is an alternating sequence (minus, plus, minus, plus, etc..). So the general term will have a (-1)^n

Also, look at the seuquence and establish a pattern. Only after you've in front of it..that will cause it to alternate. Or it will have (-1)^n in front..that will cause it to alternate.

Also, establish a pattern for the sequence. You cannot write a general term without establishing the pattern.
In this problem, the numerator increases by 1 (as it starts with 3) and the denominators are all perfect squares, starting with 16, which is 4^2.

Use the above, to try writing the general term.

Answer 2

I tried a bunch of combinations, i thought it was n/(n+1)^2 originally and then i also tried (-1)^n(n)/(n+1)^2 and my online homework still said it was wrong, so i dont know where to go from here

Answer 3

−3/16,4/25,−5/36,6/49,…

Why not just remove them? Obviously, there is a factor of "-1" in the final result. Remember that and move on with the rest.

3/16, 4/25, 5/36, 6/49,…

3/4^2
4/5^2
5/6^2
6/7^2

I'm seeing a pattern.

\(a_{n} = (-1)^{n}\dfrac{n+2}{(n+3)^{2}}\)

Are you just indexing badly?

Answer 4

so why is it n+2 on the top and n+3 on the bottom, or i guess i see why looking at it now but how did you see that from just the sequence?

Answer 5

It's just a number. Write the list and compare it.

n = 1 -- 3/4^2
n = 2 -- 4/5^2
n = 3 -- 5/6^2
n = 4 -- 6/7^2