$\begin{array}{1 1}\large\frac{n(n+1)(2n+1)}{6}+2.(2^n-1) \\ \large\frac{n(n+1)(2n+1)}{6}+2.(2^{n-1}-1) \\ \large\frac{n(n+1)(2n+1)}{6}+.(2^{n+1}-1) \\\large\frac{n(n+1)(2n+1)}{6}+.(2^n-1)\end{array} $