数学的帰納法の考え方/実例をアニメーションで理解しよう！[数学入門]

\( \displaystyle \sum_{i=1}^{n}(i) = \frac{1}{2}n(n+1) \)

\( \displaystyle \sum_{i=1}^{n}(i) = \frac{1}{2}n(n+1) \)

\( \displaystyle \sum_{i=1}^{n}(i) = \sum_{i=1}^{1}(i) = 1\)
\( \displaystyle \frac{1}{2}n(n+1) = \frac{1}{2}(1+1) = 1\)

\( \displaystyle \sum_{i=1}^{k}(i) = \frac{1}{2}k(k+1) \)

\( \displaystyle \sum_{i=1}^{k+1}(i) \)
\( = \displaystyle \sum_{i=1}^{k}(i) + (k+1) \)
↓n=kの時の式を代入
\( = \displaystyle \frac{1}{2}k(k+1) + (k+1) \)
\( = \displaystyle \frac{1}{2}\{k(k+1) + 2(k+1)\} \)
\( = \displaystyle \frac{1}{2}(k+1)(k+2) \)

\( \displaystyle \sum_{k=1}^{n}(k^2) = \frac{1}{6}n(n+1)(2n+1) \)

\( \displaystyle \sum_{i=1}^{n}(i^2) = \sum_{i=1}^{1}(i^2) = 1\)
\( \displaystyle \frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}(2 \cdot 3) = 1\)

\( \displaystyle \sum_{i=1}^{k}(i^2) = \frac{1}{6}k(k+1)(2k+1) \)

\( \displaystyle \sum_{i=1}^{k+1}(i^2) \)
\( = \displaystyle \sum_{i=1}^{k}(i^2) + (k+1)^2 \)
↓n=kの時の式を代入
\( = \displaystyle \frac{1}{6}k(k+1)(2k+1) + (k+1)^2 \)
\( = \displaystyle \frac{1}{6}\{k(k+1)(2k+1)+ 6(k+1)^2\} \)
\( = \displaystyle \frac{1}{6}\{(k+1)(2k^2+k+6k+6)\} \)
\( = \displaystyle \frac{1}{6}\{(k+1)(2k^2+7k+6)\} \)
\( = \displaystyle \frac{1}{6}(k+1)(k+2)(2k+3) \)
\( = \displaystyle \frac{1}{6}(k+1)(k+2)(2(k+1)+1) \)