Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:31Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
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3β² + 6β + 2 = 0
det| 5 9 ; 7 6 | = -33
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F = ma
∑_{k=1}^{n} k = n(n+1)/2
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det| 7 4 ; 4 8 | = 40
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\frac{8}{7} \sum_{i=1}^{n} i^{2}
det| 4 6 ; 4 5 | = -4
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