d

n \to \infty lim n \cdot (1 - Binomischer Lehrsatz (1 - \frac{1}{n})^{42}) = (5) n \to \infty lim n \cdot (1 - (\sum_{k = 0}^{42} (k 42) \cdot = 1 1^{42 - k} \cdot (- \frac{1}{n})^{k})) = n \to \infty lim n \cdot (1 - (k = 0 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k})) = (7) n \to \infty lim n \cdot (1 - ((0 42) \cdot (- \frac{1}{n})^{0} + k = 1 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k})) = n \to \infty lim n \cdot (1 - (1 + k = 1 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k})) = (7) n \to \infty lim n \cdot (1 - (1 + (1 42) \cdot (- \frac{1}{n})^{1} + k = 2 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k})) = n \to \infty lim n \cdot (1 - (1 + 42 \cdot (- \frac{1}{n}) + k = 2 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k})) = n \to \infty lim n \cdot (1 - (1 - \frac{42}{n} + k = 2 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k})) = (8) n \to \infty lim n \cdot (\frac{42}{n} + k = 2 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k}) = n \to \infty lim 42 + n \cdot k = 2 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k} = (9) n \to \infty lim 42 + lim_{n \to \infty} n \cdot k = 2 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k} = n \to \infty lim 42 + n \to \infty lim k = 2 \sum 42 (k 42) \cdot (- \frac{1}{n})^{k} \cdot n = (9) = 42 n \to \infty lim 42 + k = 2 \sum 42 = 0 lim_{n \to \infty} (k 42) \cdot (- \frac{1}{n})^{k} \cdot n = 42 + k = 2 \sum 42 0 = 42

🪴 Quartz 4.0