What's the term of $i^{n(3n-1)}$

Question

$A=i^{n(3n-1)}$

$n=0$ then $A=1$

$n=1$ then $A=-1$

$n=2$ then $A=-1$

$n=3$ then $A=1$

$n=4$ then $A=1$

So how can simplified this in term general

Answer 1

Since $i^4=1$, it is enough to consider the exponent mod $4$. We get this periodic sequence: $$ 0,2,2,0, \dots $$ Therefore, the original sequence is $$ 1,-1,-1,1, \dots $$

Answer 2

You could use Euler's formula: $$\begin{split}i^{n(3n-1)} &=\left(e^{\pi i/2}\right)^{n(3n-1)}\\&=e^{\frac{1}{2}\pi in(3n-1)} \\&=\cos\left(\frac{1}{2}\pi n(3n-1)\right)+i\sin\left(\frac{1}{2}\pi n(3n-1)\right)\end{split}$$ and note that $(3n-1)-n=2n-1$ is odd, so one of $n$, $3n-1$ is even so $$\frac{1}{2}\pi n(3n-1)$$ will always be an integer multiple of $\pi$ so taking the sine of it always yields $0$, so $$i^{n(3n-1)}=\cos\left(\frac{1}{2}\pi n(3n-1)\right).$$ Within a cosine, we can always subtract an even multiple of $\pi$ without changing it's value. Note that $2n^2\pi$ is an even multiple of $\pi$, so we can say that $\cos\left(\frac{1}{2}\pi n(3n-1)\right)=\cos\left(\frac{1}{2}\pi n(-n-1)\right)=\cos\left(\frac{1}{2}\pi n(n+1)\right)$ where the final equality follows from the identity $\cos x=\cos(-x)$. Now, we see that if $n\equiv 0\mod 4$ or $n\equiv -1\mod 4$, we take the cosine of an even multiple of $\pi$ which yields $1$. If $n\equiv 1\mod 4$ or $n\equiv 2\mod 4$, then we take the cosine of an odd multiple of $\pi$ which yields $-1$. In the end, we get $$i^{n(3n-1)}=\begin{cases}1,&n\equiv0\mod4\\-1,&n\equiv1\mod4\\-1,&n\equiv2\mod4\\1,&n\equiv3\mod4\end{cases}$$ so the repeated pattern is $1,-1,-1,1,\ldots$.