Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:27Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
276K
F = ma
det| 9 4 ; 1 10 | = 86
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\frac{8}{7} \sum_{i=1}^{n} i^{2}
∑_{k=1}^{n} k = n(n+1)/2
1K
∫₀^∞ e^(-x²) dx = √π / 2
153K
∑_{k=1}^{n} k = n(n+1)/2
338K