Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:40Z
∑_{k=1}^{n} k = n(n+1)/2
8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
129K
∑_{k=1}^{n} k = n(n+1)/2
det| 9 4 ; 1 10 | = 86
390K
det| 12 1 ; 4 13 | = 152
1K
∑_{k=1}^{n} k = n(n+1)/2
153K