高専の数学解説

第2章 2.12

(1)

Sn=∑n=12n−1n2−∑n=1n−1n2=(2n−1)2n(4n−1)6−(n−1)n(2n−1)6=n6{2(2n−1)(4n−1)−(n−1)(2n−1)}S_n = \sum_{n=1}^{2n-1} n^2 - \sum_{n=1}^{n-1} n^2 = \frac{(2n-1)2n(4n-1)}{6} - \frac{(n-1)n(2n-1)}{6} = \frac{n}{6}\{2(2n-1)(4n-1) - (n-1)(2n-1)\}

=n6(16n2−12n+2−2n2+3n−1)=n(2n−1)(7n−1)6= \frac{n}{6}(16n^2 - 12n + 2 - 2n^2 + 3n - 1) = \frac{n(2n-1)(7n-1)}{6}

(2)

lim⁡n→∞Snn3=lim⁡n→∞n(2n−1)(7n−1)6n3=lim⁡n→∞n(2n−1)(7n−1)6n3=lim⁡n→∞(2−1n)(7−1n)6=73\lim_{n \to \infty} \frac{S_n}{n^3} = \lim_{n \to \infty} \frac{\dfrac{n(2n-1)(7n-1)}{6}}{n^3} = \lim_{n \to \infty} \frac{n(2n-1)(7n-1)}{6n^3} = \lim_{n \to \infty} \frac{\left(2 - \dfrac{1}{n}\right)\left(7 - \dfrac{1}{n}\right)}{6} = \frac{7}{3}

解説: r31bn1z