Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:35Z
∑_{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| 1 1 ; 5 2 | = -3
276K
F = ma
det| 4 6 ; 4 5 | = -4
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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| 5 6 ; 2 6 | = 18
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det| 9 4 ; 1 10 | = 86
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