Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:59Z
∑_{k=1}^{n} k = n(n+1)/2
∫₀^∞ e^(-z²) dz = √π / 2
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8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
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∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
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\lim_{r\to 0} \frac{\sin r}{r} = 1
det| 9 4 ; 1 10 | = 86
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det| 3 6 ; 0 4 | = 12
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∑_{k=1}^{n} k = n(n+1)/2
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det| 7 9 ; 1 8 | = 47
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