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

Aug 1, 2018

${T}_{n} = {\left(- 1\right)}^{n + 1} {n}^{2}$

#### Explanation:

The formula for this sequence is

${T}_{n} = {\left(- 1\right)}^{n + 1} {n}^{2}$
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 ${\left(- 1\right)}^{2}$, you will always get a positive $1$
• If you imagine cubing $- 1$ ie ${\left(- 1\right)}^{3}$, you will always get a negative $1$
• Hence, if you write ${\left(- 1\right)}^{\text{even number} + 1}$, you will get a negative $1$ ie ${\left(- 1\right)}^{2 + 1} = {\left(- 1\right)}^{3} = - 1$ ; ${\left(- 1\right)}^{4 + 1} = {\left(- 1\right)}^{5} = - 1$
• Also, if you write ${\left(- 1\right)}^{\text{odd number} + 1}$, you will get a positive $1$ ie ${\left(- 1\right)}^{1 + 1} = {\left(- 1\right)}^{2} = 1$ ; ${\left(- 1\right)}^{3 + 1} = {\left(- 1\right)}^{4} = 1$

Thus, ${T}_{n} = {\left(- 1\right)}^{n + 1} {n}^{2}$