Question

23 1997 明治大
\\\sum_{k=1}^{400} \frac{1}{\sqrt{k} + \sqrt{k+1}} \text{を求めよ。}\\

Explanation:

分母の有理化

$$ \frac{1}{\sqrt{k} + \sqrt{k+1}} = \frac{\sqrt{k+1} - \sqrt{k}}{(\sqrt{k+1} + \sqrt{k})(\sqrt{k+1} - \sqrt{k})} = \sqrt{k+1} - \sqrt{k} $$

和の計算（相殺）

$$ \sum_{k=1}^{400} \frac{1}{\sqrt{k} + \sqrt{k+1}} = \sum_{k=1}^{400} (\sqrt{k+1} - \sqrt{k}) $$

$$ = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \dots + (\sqrt{401} - \sqrt{400}) $$

$$ = \sqrt{401} - 1 $$

Answer:

$$\sqrt{401} - 1$$