Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:03Z
∑_{k=1}^{n} k = n(n+1)/2
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
\lim_{r\to 0} \frac{\sin r}{r} = 1
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
153K
det| 12 1 ; 1 13 | = 155
∫₀^∞ e^(-φ²) dφ = √π / 2
338K