Question

18．已知數(shù)列{a_n}的通項(xiàng)公式為a_n=$\frac{1}{n^2+3n}$，求數(shù)列{a_n}前n項(xiàng)和S_n．

Answer 1

分析 由a_n=$\frac{1}{n^2+3n}$=$\frac{1}{3}（\frac{1}{n}-\frac{1}{n+3}）$，利用“裂項(xiàng)求和”即可得出．

解答 解：∵a_n=$\frac{1}{n^2+3n}$=$\frac{1}{3}（\frac{1}{n}-\frac{1}{n+3}）$，
∴數(shù)列{a_n}前n項(xiàng)和S_n=$\frac{1}{3}[（1-\frac{1}{4}）+（\frac{1}{2}-\frac{1}{5}）+（\frac{1}{3}-\frac{1}{6}）+（\frac{1}{4}-\frac{1}{7}）$+…+$（\frac{1}{n-3}-\frac{1}{n}）$+$（\frac{1}{n-2}-\frac{1}{n+1}）$+$（\frac{1}{n-1}-\frac{1}{n+2}）$+$（\frac{1}{n}-\frac{1}{n+3}）]$
=$\frac{1}{3}（1+\frac{1}{2}+\frac{1}{3}$-$\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}）$
=$\frac{1}{3}（\frac{11}{6}$-$\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}）$

點(diǎn)評 本題考查了數(shù)列的“裂項(xiàng)求和”方法、變形能力，考查了推理能力與計(jì)算能力，屬于中檔題．