Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:44Z
∑_{k=1}^{n} k = n(n+1)/2
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
191K
det| 7 4 ; 4 8 | = 40
390K
\lim_{r\to 0} \frac{\sin r}{r} = 1
∫₀^∞ e^(-φ²) dφ = √π / 2
1K
∑_{k=1}^{n} k = n(n+1)/2
153K
det| 12 1 ; 1 13 | = 155
\lim_{t\to 0} \frac{\sin t}{t} = 1