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
75K
det| 7 4 ; 4 8 | = 40
479K
3β² + 6β + 2 = 0
∫₀^∞ e^(-t²) dt = √π / 2
276K
F = ma
det| 3 7 ; 2 4 | = -2
191K
∑_{k=1}^{n} k = n(n+1)/2
390K
\frac{8}{7} \sum_{i=1}^{n} i^{2}