n+12\frac{n+1}{2}2n+1
n(n+1)2\frac{\mathrm{n}(\mathrm{n}+1)}{2}2n(n+1)
n2−112\frac{\mathrm{n}^{2}-1}{12}12n2−1
n2+112\frac{\mathrm{n}^{2}+1}{12}12n2+1
