Question

8．已知數(shù)列{a_n}中，${a_n}=\frac{1}{{（\sqrt{n-1}+\sqrt{n}）（\sqrt{n-1}+\sqrt{n+1}）（\sqrt{n}+\sqrt{n+1}}}$，則S₄=$\frac{3-\sqrt{5}}{2}$．

Answer 1

分析 ${a_n}=\frac{1}{{（\sqrt{n-1}+\sqrt{n}）（\sqrt{n-1}+\sqrt{n+1}）（\sqrt{n}+\sqrt{n+1}}}$=$\frac{1}{2}（\frac{1}{\sqrt{n-1}+\sqrt{n}}-\frac{1}{\sqrt{n}+\sqrt{n+1}}）$．利用裂項(xiàng)求和方法即可得出．

解答 解：${a_n}=\frac{1}{{（\sqrt{n-1}+\sqrt{n}）（\sqrt{n-1}+\sqrt{n+1}）（\sqrt{n}+\sqrt{n+1}}}$=$\frac{1}{2}（\frac{1}{\sqrt{n-1}+\sqrt{n}}-\frac{1}{\sqrt{n}+\sqrt{n+1}}）$．
則S₄=$\frac{1}{2}[（1-\frac{1}{1+\sqrt{2}}）$+$（\frac{1}{1+\sqrt{2}}-\frac{1}{\sqrt{2}+\sqrt{3}}）$+$（\frac{1}{\sqrt{2}+\sqrt{3}}-\frac{1}{\sqrt{3}+2}）$+$（\frac{1}{\sqrt{3}+2}-\frac{1}{2+\sqrt{5}}）]$
=$\frac{1}{2}（1-\frac{1}{2+\sqrt{5}}）$
=$\frac{3-\sqrt{5}}{2}$．
故答案為：$\frac{3-\sqrt{5}}{2}$．

點(diǎn)評(píng) 本題考查了裂項(xiàng)求和方法法、分母有理化，考查了推理能力與計(jì)算能力，屬于中檔題．