次の和 $S$ を求める問題です。 $S = \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{n} + \sqrt{n+1}}$解析学級数有理化telescoping sum根号2025/6/261. 問題の内容次の和 SSS を求める問題です。S=11+2+12+3+13+4+⋯+1n+n+1S = \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{n} + \sqrt{n+1}}S=1+21+2+31+3+41+⋯+n+n+112. 解き方の手順各項の分母を有理化します。1k+k+1=k−k+1(k+k+1)(k−k+1)=k−k+1k−(k+1)=k−k+1−1=k+1−k\frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{\sqrt{k} - \sqrt{k+1}}{(\sqrt{k} + \sqrt{k+1})(\sqrt{k} - \sqrt{k+1})} = \frac{\sqrt{k} - \sqrt{k+1}}{k - (k+1)} = \frac{\sqrt{k} - \sqrt{k+1}}{-1} = \sqrt{k+1} - \sqrt{k}k+k+11=(k+k+1)(k−k+1)k−k+1=k−(k+1)k−k+1=−1k−k+1=k+1−kしたがって、S=(2−1)+(3−2)+(4−3)+⋯+(n+1−n)S = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \cdots + (\sqrt{n+1} - \sqrt{n})S=(2−1)+(3−2)+(4−3)+⋯+(n+1−n)これはtelescoping sum(隣り合う項同士で打ち消しあう和)です。S=−1+(2−2)+(3−3)+⋯+(n−n)+n+1S = -\sqrt{1} + (\sqrt{2} - \sqrt{2}) + (\sqrt{3} - \sqrt{3}) + \cdots + (\sqrt{n} - \sqrt{n}) + \sqrt{n+1}S=−1+(2−2)+(3−3)+⋯+(n−n)+n+1S=−1+n+1S = -1 + \sqrt{n+1}S=−1+n+1S=n+1−1S = \sqrt{n+1} - 1S=n+1−13. 最終的な答えS=n+1−1S = \sqrt{n+1} - 1S=n+1−1
