1.

\sqrt{\frac{\sqrt{n + k}+\sqrt{k}}{\sqrt{n + k}+\sqrt{n}}}=\frac{\sqrt{k}+\sqrt{n + k}-\sqrt{n}}{\sqrt{2k}}
\sqrt{\frac{\sqrt{n + k}+\sqrt{k}}{\sqrt{n + k}+\sqrt{n}}}=\frac{\sqrt{k}+\sqrt{n + k}-\sqrt{n}}{\sqrt{2k}}
（2025圆梦14？）

有良好的裂项性质，取k=1,2适合出求和或者放缩题

当

k = 2
k = 2
时，
a_n = \frac{\sqrt{2} + \sqrt{n + 2} - \sqrt{n}}{2}\\
a_n = \frac{\sqrt{2} + \sqrt{n + 2} - \sqrt{n}}{2}\\
求前
n
n
项和
S_n = \sum_{i=1}^{n} a_i
S_n = \sum_{i=1}^{n} a_i

S_n = \frac{1}{2} \sum_{i=1}^{n} (\sqrt{2} + \sqrt{i + 2} - \sqrt{i}) \\ = \frac{1}{2} \left( n\sqrt{2} + \sum_{i=1}^{n} (\sqrt{i + 2} - \sqrt{i}) \right)\\ \\
S_n = \frac{1}{2} \sum_{i=1}^{n} (\sqrt{2} + \sqrt{i + 2} - \sqrt{i}) \\ = \frac{1}{2} \left( n\sqrt{2} + \sum_{i=1}^{n} (\sqrt{i + 2} - \sqrt{i}) \right)\\ \\

对

\sum_{i=1}^{n} (\sqrt{i + 2} - \sqrt{i})
\sum_{i=1}^{n} (\sqrt{i + 2} - \sqrt{i})
进行裂项相消：

\sum_{i=1}^{n} (\sqrt{i + 2} - \sqrt{i}) = (\sqrt{3} - 1) + (2 - \sqrt{2}) + (\sqrt{5} - \sqrt{3}) + \cdots + (\sqrt{n + 2} - \sqrt{n}) \\ = \sqrt{n + 1} + \sqrt{n + 2} - 1 - \sqrt{2}\\ \\
\sum_{i=1}^{n} (\sqrt{i + 2} - \sqrt{i}) = (\sqrt{3} - 1) + (2 - \sqrt{2}) + (\sqrt{5} - \sqrt{3}) + \cdots + (\sqrt{n + 2} - \sqrt{n}) \\ = \sqrt{n + 1} + \sqrt{n + 2} - 1 - \sqrt{2}\\ \\
代入
S_n
S_n
：

S_n = \frac{1}{2} \left( n\sqrt{2} + \sqrt{n + 1} + \sqrt{n + 2} - 1 - \sqrt{2} \right) \\ = \frac{\sqrt{n + 1} + \sqrt{n + 2} + (n - 1)\sqrt{2} - 1}{2} \\
S_n = \frac{1}{2} \left( n\sqrt{2} + \sqrt{n + 1} + \sqrt{n + 2} - 1 - \sqrt{2} \right) \\ = \frac{\sqrt{n + 1} + \sqrt{n + 2} + (n - 1)\sqrt{2} - 1}{2} \\

2.

\(\sqrt{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}+\dfrac{1}{(a + b)^{2}}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a + b}\)
\(\sqrt{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}+\dfrac{1}{(a + b)^{2}}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a + b}\)

也具有良好的性质，其中取

b
b
为常数，令
a=n
a=n
，构成典型的裂项形式，可以出求和或者放缩题

……

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