# How do I find the nth term of this sequence?

#### Raanikeri

##### New member

Could it be some sort of geometric sequence? Or something involving binomial expansion?
There is no common difference at any level, it seems, so I don't think it's an arithmetic or quadratic sequence.

There is a pattern in the sequence, and I could easily predict what the next few terms will be (-6, 7, -8...), but I am stumped when it comes to finding the nth term.

Can anyone help?

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Could it be some sort of geometric sequence? Or something involving binomial expansion?
There is no common difference at any level, it seems, so I don't think it's an arithmetic or quadratic sequence.

There is a pattern in the sequence, and I could easily predict what the next few terms will be (-6, 7, -8...), but I am stumped when it comes to finding the nth term.

Can anyone help?
The nth term would be [imath](-1)^{n+1}n[/imath].

-Dan

Hint: use [imath](-1)^{n+1}[/imath]

[imath]a_n = (-1)^{n+1}b_n[/imath]

It is easy to guess [imath]b_n[/imath] now.

Forget the - signs for a moment. Then the sequence is 1 2 3 4 5... Clearly the nth term in my sequence will be n. You get to show how you would decide if the n is + or -.

If $$\displaystyle n$$ is odd, then $$\displaystyle t_n = n$$.

If $$\displaystyle n$$ is even, then $$\displaystyle t_n = -n$$.

To combine these statements, we need a way of producing +1 if n is odd and -1 if n is even.

One way of doing this is looking at odd/even powers of -1.

Note that (-1)^even is +1 and (-1)^odd is -1.

So, if n is odd, we need a coefficient of +1, so the index must be even. Note that (n+1) is even, since n is odd.

Similarly, if n is even, we need a coefficient of -1, so the index must be odd. Note that, in this case, (n+1) is odd, since n is even.

So, for both cases, $$\displaystyle t_n = (-1)^{n+1}n$$ .

An index of $$\displaystyle (n-1), (n+3), (n+5)$$ etc would also work, but let's keep it simple.