Question

4．已知數(shù)列{a_n}的前n項(xiàng)和為S_n，a_n=$\frac{1}{（\sqrt{n-1}+\sqrt{n}）（\sqrt{n-1}+\sqrt{n+1}）（\sqrt{n}+\sqrt{n+1}）}$，則S₂₀₁₆=$\frac{1+12\sqrt{14}-\sqrt{2017}}{2}$．

Answer 1

分析 將a_n=$\frac{1}{（\sqrt{n-1}+\sqrt{n}）（\sqrt{n-1}+\sqrt{n+1}）（\sqrt{n}+\sqrt{n+1}）}$分子分母同乘$\sqrt{n+1}-\sqrt{n-1}$，再使用裂項(xiàng)法得出a_n=$\frac{1}{2}$（$\frac{1}{\sqrt{n-1}+\sqrt{n}}$-$\frac{1}{\sqrt{n}+\sqrt{n+1}}$），從而得出S₂₀₁₆的值．

解答 解：a_n=$\frac{1}{（\sqrt{n-1}+\sqrt{n}）（\sqrt{n-1}+\sqrt{n+1}）（\sqrt{n}+\sqrt{n+1}）}$=$\frac{\sqrt{n+1}-\sqrt{n-1}}{2（\sqrt{n-1}+\sqrt{n}）（\sqrt{n}+\sqrt{n+1}）}$
=$\frac{1}{2}$（$\frac{1}{\sqrt{n-1}+\sqrt{n}}$-$\frac{1}{\sqrt{n}+\sqrt{n+1}}$）．
∴S₂₀₁₆=$\frac{1}{2}$（1-$\frac{1}{1+\sqrt{2}}$）+$\frac{1}{2}$（$\frac{1}{1+\sqrt{2}}$-$\frac{1}{\sqrt{2}+\sqrt{3}}$）+$\frac{1}{2}$（$\frac{1}{\sqrt{2}+\sqrt{3}}$-$\frac{1}{\sqrt{3}+2}$）
+…+$\frac{1}{2}$（$\frac{1}{\sqrt{2015}+\sqrt{2016}}$-$\frac{1}{\sqrt{2016}+\sqrt{2017}}$）
=$\frac{1}{2}$（1-$\frac{1}{1+\sqrt{2}}$+$\frac{1}{1+\sqrt{2}}$-$\frac{1}{\sqrt{2}+\sqrt{3}}$+$\frac{1}{\sqrt{2}+\sqrt{3}}$-$\frac{1}{\sqrt{3}+2}$+…+$\frac{1}{\sqrt{2015}+\sqrt{2016}}$-$\frac{1}{\sqrt{2016}+\sqrt{2017}}$）
=$\frac{1}{2}$（1-$\frac{1}{\sqrt{2016}+\sqrt{2017}}$）
=$\frac{1}{2}$[1-（$\sqrt{2017}-\sqrt{2016}$）]
=$\frac{1+\sqrt{2016}-\sqrt{2017}}{2}$
=$\frac{1+12\sqrt{14}-\sqrt{2017}}{2}$．
故答案為：$\frac{1+12\sqrt{14}-\sqrt{2017}}{2}$．

點(diǎn)評(píng) 本題考查了裂項(xiàng)法數(shù)列求和，尋找通項(xiàng)公式的分母特點(diǎn)進(jìn)行裂項(xiàng)是關(guān)鍵，屬于中檔題．