để 20\(⋮2n+1\)
=> \(\dfrac{20}{2n+1}\) phải nguyên thì
2n+1\(\inƯ\left(20\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm10;\pm20\right\}\)
ta có bảng sau
2n+1 | -20 | -10 | -5 | -4 | -2 | -1 | 1 | 2 | 4 | 5 | 10 | 20 |
2n | -21 | -11 | -6 | -5 | -3 | -2 | 0 | 1 | 3 | 4 | 9 | 19 |
n | \(-\dfrac{21}{2}\) | \(-\dfrac{11}{2}\) | -3 | -\(\dfrac{5}{2}\) | -\(\dfrac{3}{2}\) | -1 | 0 | \(\dfrac{1}{2}\) | \(\dfrac{3}{2}\) | 2 | \(\dfrac{9}{2}\) | \(\dfrac{19}{2}\) |
vậy .....