Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:23Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
F = ma
∑_{k=1}^{n} k = n(n+1)/2
191K
det| 11 4 ; 5 12 | = 112
390K
\frac{8}{7} \sum_{i=1}^{n} i^{2}
∑_{k=1}^{n} k = n(n+1)/2
153K
det| 7 4 ; 4 8 | = 40
338K
det| 8 9 ; 4 9 | = 36
188K