Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:41Z
∑_{k=1}^{n} k = n(n+1)/2
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
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∑_{k=1}^{n} k = n(n+1)/2
det| 1 4 ; 3 2 | = -10
191K
det| 11 4 ; 5 12 | = 112
390K
det| 9 4 ; 1 10 | = 86
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∫₀^∞ e^(-t²) dt = √π / 2
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∑_{k=1}^{n} k = n(n+1)/2
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