Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:03Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
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3β² + 6β + 2 = 0
F = ma
∑_{k=1}^{n} k = n(n+1)/2
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\frac{8}{7} \sum_{i=1}^{n} i^{2}
∑_{k=1}^{n} k = n(n+1)/2
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det| 8 9 ; 4 9 | = 36
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