Unter der jeweiligen Induktionsannahme gilt

\( \sum_{k=1}^{n+1} k \cdot 2^k \) \(= (n+1) \cdot 2^{n+1} + \sum_{k=1}^n k \cdot 2^k \) \(= (n+1) \cdot 2^{n+1} + 2 + 2^{n+1} \cdot (n-1) = 2 + 2^{n+1} \cdot (n+1+n-1) \) \(= 2 + 2^{n+1} \cdot 2n \) \(= 2 + 2^{(n+1)+1} \cdot ((n+1)-1) \)

und

\( \sum_{k=1}^{n+1} (2k-1)^2 \) \(= (2(n+1)-1)^2 + \sum_{k=1}^n (2k-1)^2 \) \(= (2n+1)^2 + \frac{1}{3}n(2n+1)(2n-1) \) \(= (2n+1+\frac{1}{3}n(2n-1))(2n+1) \) \(= \frac{1}{3} (6n+3+n(2n-1))(2n+1) \) \(= \frac{1}{3} (2n^2+5n+3)(2n+1) \) \(= \frac{1}{3}(n+1)(2n+3)(2n+1) \) \(= \frac{1}{3}(n+1)(2(n+1)+1)(2(n+1)-1) \)