\(\dfrac{4n+7}{2n+1}=\dfrac{\left(4n+2\right)+5}{2n+1}=\dfrac{2\left(2n+1\right)+5}{2n+1}=2+\dfrac{5}{2n+1}\)
Để \(\dfrac{4n+7}{2n+1}\) nguyên thì \(\dfrac{5}{2n+1}\) nguyên \(\Rightarrow2n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng:
2n+1 | -5 | -1 | 1 | 5 |
n | -3 | -1 | 0 | 2 |
\(S=-3+\left(-1\right)+0+2=-2\)
Chọn B