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Melody

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Melody  11 февр. 2022 г.
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14 февр. 2022 г.
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Thanks Tiggsy

I can follow all that.  Similar to mine just better :))

 

z11z2=14{1sin2(π/14)+1sin2(3π/14)++1sin2(13π/14)}. z11z2=14{2sin2(π/14)+2sin2(3π/14)+2sin2(5π/14)+1sin2(π/2)}. z11z2=14{2sin2(π/14)+2sin2(3π/14)+2sin2(5π/14)+1}. z11z2=14+12{1sin2(π/14)+1sin2(3π/14)+1sin2(5π/14)}. z11z2=14+12{24}. z11z2=12.25

 

 

 

 

 

 

LaTex

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\dots +\frac{1}{\sin^{2}(13\pi/14)}\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+\frac{1}{\sin^{2}(\pi/2)}
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}\left\{\frac{2}{\sin^{2}(\pi/14)} +\frac{2}{\sin^{2}(3\pi/14)}+\frac{2}{\sin^{2}(5\pi/14)}+1
\right\}.\\~\\
\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{\frac{1}{\sin^{2}(\pi/14)} +\frac{1}{\sin^{2}(3\pi/14)}+\frac{1}{\sin^{2}(5\pi/14)}
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}
=\frac{1}{4}+\frac{1}{2}\left\{24
\right\}.\\~\\

\displaystyle \sum_{z}\frac{1}{\mid 1-z\mid^{2}}=12.25

13 февр. 2022 г.