用裂项相消法证明
∑r=1N1r+r+1=N+1−1.\sum_{r=1}^{N} \frac{1}{\sqrt{r} + \sqrt{r+1}} = \sqrt{N+1} - 1.

数列与级数概念通项公式English
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完整解法

3(i)

1. Rationalise the denominator of the general term.

1r+r+1=1r+1+r×r+1−rr+1−r\frac{1}{\sqrt{r} + \sqrt{r+1}} = \frac{1}{\sqrt{r+1} + \sqrt{r}} \times \frac{\sqrt{r+1} - \sqrt{r}}{\sqrt{r+1} - \sqrt{r}}
=r+1−r(r+1)−r=r+1−r= \frac{\sqrt{r+1} - \sqrt{r}}{(r+1) - r} = \sqrt{r+1} - \sqrt{r}

2. Expand the sum to list the first few and last few terms, identifying the telescoping pattern.

∑r=1N(r+1−r)\sum_{r=1}^{N} (\sqrt{r+1} - \sqrt{r})
Evaluating the terms from r=1r = 1 to r=Nr = N:
When r=1r=1: 2−1\sqrt{2} - \sqrt{1}
When r=2r=2: 3−2\sqrt{3} - \sqrt{2}
When r=3r=3: 4−3\sqrt{4} - \sqrt{3}
…\dots
When r=N−1r=N-1: N−N−1\sqrt{N} - \sqrt{N-1}
When r=Nr=N: N+1−N\sqrt{N+1} - \sqrt{N}

3. Cancel the intermediate terms to find the final simplified sum.

Summing the terms, all intermediate square roots cancel out:
=N+1−1= \sqrt{N+1} - \sqrt{1}
=N+1−1= \sqrt{N+1} - 1

答案

\sqrt{N+1} - 1 \text{ (shown)}

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