Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:22Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
276K
det| 5 9 ; 7 6 | = -33
129K
F = ma
∑_{k=1}^{n} k = n(n+1)/2
191K
det| 7 4 ; 4 8 | = 40
390K
\frac{8}{7} \sum_{i=1}^{n} i^{2}