Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:16Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
75K
3β² + 6β + 2 = 0
F = ma
∑_{k=1}^{n} k = n(n+1)/2
191K
\frac{8}{7} \sum_{i=1}^{n} i^{2}
1x² + 5x + 7 = 0
∑_{k=1}^{n} k = n(n+1)/2
473K