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

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Answer 1 · 2018-08-01T08:07:07

$T_{n} = {(- 1)}^{n + 1} n^{2}$ $T_n=(-1)^(n+1)n^2$

The formula for this sequence is

$T_{n} = {(- 1)}^{n + 1} n^{2}$ $T_n=(-1)^(n+1)n^2$
where $n$ $n$ 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 $2^{2} = 4$ $2^2=4$ , $3^{2} = 9$ $3^2=9$ and $4^{2} = 16$ $4^2=16$
notice that if you consider $1$ $1$ as your 1st term, then every even term is negative ie $1$ $1$ is your 1st term, $- 4$ $-4$ is your 2nd term, $9$ $9$ is your 3rd term and $- 16$ $-16$ is your 4th term
If you imagine squaring $- 1$ $-1$ ie ${(- 1)}^{2}$ $(-1)^2$ , you will always get a positive $1$ $1$
If you imagine cubing $- 1$ $-1$ ie ${(- 1)}^{3}$ $(-1)^3$ , you will always get a negative $1$ $1$
Hence, if you write ${(- 1)}^{even number + 1}$ $(-1)^("even number"+1)$ , you will get a negative $1$ $1$ ie ${(- 1)}^{2 + 1} = {(- 1)}^{3} = - 1$ $(-1)^(2+1)=(-1)^3=-1$ ; ${(- 1)}^{4 + 1} = {(- 1)}^{5} = - 1$ $(-1)^(4+1)=(-1)^5=-1$
Also, if you write ${(- 1)}^{odd number + 1}$ $(-1)^("odd number"+1)$ , you will get a positive $1$ $1$ ie ${(- 1)}^{1 + 1} = {(- 1)}^{2} = 1$ $(-1)^(1+1)=(-1)^2=1$ ; ${(- 1)}^{3 + 1} = {(- 1)}^{4} = 1$ $(-1)^(3+1)=(-1)^4=1$

Thus, $T_{n} = {(- 1)}^{n + 1} n^{2}$ $T_n=(-1)^(n+1)n^2$