I want to show following equation holds. \begin{align} (-1)^{\frac{(n-1)(n-2)}{2}} = -\frac{1}{2} [ (1+i) i^n + (1-i)(-i)^n] \end{align} Can you give some hint or any practical method to compute this?
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Consider the sequence $$ a_n=hr^n+ks^n $$ with $r\ne s$. The sequence satisfies the recursion $$ a_{n+2}-(r+s)a_{n+1}+rsa_n=0 $$ with initial values $$ a_0=h+k,\quad a_1=hr+ks $$ In the case where $r=i$ and $s=-i$, the recursion is $a_{n+2}=-a_n$. The initial values, for $h=(1+i)/2$ and $k=(1-i)/2$ are $$ a_0=1,\quad a_1=-1 $$ So your sequence is $$ 1,-1,-1,1,1,-1,-1,1,1,\dotsc $$ and $a_n=(-1)^{(n-1)(n-2)/2}=(-1)^{n(n+1)/2}$ satisfies the requirement.
egreg
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1I think that this is the best perspective here. – Martin Brandenburg Dec 15 '15 at 11:25
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Hint: Consider the real part of $(1+i)i^n$.
Solution:
The LHS is $-Re((1+i)i^n)$. Now $i^n=1,i,-1,-i,\dots$ is a periodic sequence of period $4$. So, $-Re((1+i)i^n)$ is $-1,1,1,-1,\dots$ is also a periodic sequence of period $4$. The RHS is that same sequence.
lhf
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1I'd like to see a solution where $(-1)^{\frac{(n-1)(n-2)}{2}}$ appears naturally. – lhf Dec 15 '15 at 10:56
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There are actually four possibilities:
If $n$ is of the form $4k+1$ of $4k+2$ then it is $1=1$. If $n$ is of the form $4k+3$ of $4k+0$ then it is $-1=-1$.
vudu vucu
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