Question

Find the Nth term of the following sequence?: 1, -4, 9, -16, ...

Answer 1

`T_n=(-1)^(n+1)n^2`

Explanation:

The formula for this sequence is

`T_n=(-1)^(n+1)n^2`
where `n` is the nth term

How did I get this formula?

• If you look at the numbers, you will notice that they are all square numbers ie `2^2=4`, `3^2=9` and `4^2=16`
• notice that if you consider `1` as your 1st term, then every even term is negative ie `1` is your 1st term, `-4` is your 2nd term, `9` is your 3rd term and `-16` is your 4th term
• If you imagine squaring `-1` ie `(-1)^2`, you will always get a positive `1`
• If you imagine cubing `-1` ie `(-1)^3`, you will always get a negative `1`
• Hence, if you write `(-1)^("even number"+1)`, you will get a negative `1` ie `(-1)^(2+1)=(-1)^3=-1` ; `(-1)^(4+1)=(-1)^5=-1`
• Also, if you write `(-1)^("odd number"+1)`, you will get a positive `1` ie `(-1)^(1+1)=(-1)^2=1` ; `(-1)^(3+1)=(-1)^4=1`

Thus, `T_n=(-1)^(n+1)n^2`