Find the sum of the 50 th powers of all sides and diagonals…

Find the sum of the $50$th powers of all sides and diagonals of a regular $100$-gon inscribed in a circle of radius $R$.

Step 1 · Chord lengths and their multiplicities

With $n=100$, the chord between vertices separated by $m$ positions has length $d_m=2R\sin\frac{m\pi}{100}$. For $m=1,\ldots,49$, there are exactly $100$ pairs at that separation; for $m=50$ (diametrically opposite pairs), there are exactly $50$ pairs. Total: $49(100)+50=4950=\binom{100}2$ ✓.

Step 2 · Set up the sum of $50$th powers

$$ S=\sum_{m=1}^{49}100\,d_m^{50}+50\,d_{50}^{50}=100(2R)^{50}\sum_{m=1}^{49}\sin^{50}\frac{m\pi}{100}+50(2R)^{50}\sin^{50}\frac\pi2 $$

Since $\sin\frac\pi2=1$:

$$ S=100\cdot2^{50}R^{50}\sum_{m=1}^{49}\sin^{50}\frac{m\pi}{100}+50\cdot2^{50}R^{50} $$

Step 3 · The classical sine-power-sum identity

For a positive even power $2p$: $\ \displaystyle\sum_{m=1}^{n-1}\sin^{2p}\frac{m\pi}n=\frac n{2^{2p}}\binom{2p}p\ $ (a consequence of the binomial expansion of $\sin^{2p}\theta$ via Euler's formula, summed over the $n$-th roots of unity). With $n=100$, $2p=50$:

$$ \sum_{m=1}^{99}\sin^{50}\frac{m\pi}{100}=\frac{100}{2^{50}}\binom{50}{25} $$

Step 4 · Isolate the sum over $m=1,\ldots,49$ using symmetry

Since $\sin\frac{m\pi}{100}=\sin\frac{(100-m)\pi}{100}$, the terms $m=1,\ldots,49$ mirror $m=51,\ldots,99$, and $m=50$ contributes $\sin^{50}\frac\pi2=1$ on its own:

$$ \sum_{m=1}^{99}\sin^{50}\frac{m\pi}{100}=2\sum_{m=1}^{49}\sin^{50}\frac{m\pi}{100}+1 \quad\Longrightarrow\quad \sum_{m=1}^{49}\sin^{50}\frac{m\pi}{100}=\frac12\left(\frac{100}{2^{50}}\binom{50}{25}-1\right) $$

Step 5 · Substitute back and simplify

$$ S=100\cdot2^{50}R^{50}\cdot\frac12\left(\frac{100}{2^{50}}\binom{50}{25}-1\right)+50\cdot2^{50}R^{50} $$

$$ =50R^{50}\left(100\binom{50}{25}-2^{50}\right)+50\cdot2^{50}R^{50}=5000R^{50}\binom{50}{25}-50\cdot2^{50}R^{50}+50\cdot2^{50}R^{50} $$

The $2^{50}$ terms cancel exactly:

$$ \boxed{S=5000R^{50}\binom{50}{25}} $$

Verification

Direct brute-force computation (summing $|A_i-A_j|^{50}$ over all $4950$ vertex pairs of an actual regular $100$-gon, $R=1$) gives $S\approx6.32053032188756\times10^{17}$, matching $5000\binom{50}{25}=5000\times126{,}410{,}606{,}437{,}752=632{,}053{,}032{,}188{,}760{,}000=6.3205303218876\times10^{17}$ to full floating-point precision.