Question

17．已知數(shù)列{a_n}的通項(xiàng)a_n=$\frac{2}{4{n}^{2}-4n-3}$，則其前n項(xiàng)和為-$\frac{2n}{4{n}^{2}-1}$．

Answer 1

分析 由于a_n=$\frac{2}{4{n}^{2}-4n-3}$=$\frac{1}{2}$$（\frac{1}{2n-3}-\frac{1}{2n+1}）$，利用“裂項(xiàng)求和”即可得出．

解答 解：∵a_n=$\frac{2}{4{n}^{2}-4n-3}$=$\frac{1}{2}$$（\frac{1}{2n-3}-\frac{1}{2n+1}）$，
∴其前n項(xiàng)和=$\frac{1}{2}$$[（-1-\frac{1}{3}）$+$（1-\frac{1}{5}）$+$（\frac{1}{3}-\frac{1}{7}）$+…+$（\frac{1}{2n-5}-\frac{1}{2n-1}）$+$（\frac{1}{2n-3}-\frac{1}{2n+1}）]$
=$\frac{1}{2}（-1+1-\frac{1}{2n-1}-\frac{1}{2n+1}）$
=-$\frac{2n}{4{n}^{2}-1}$．
故答案為：-$\frac{2n}{4{n}^{2}-1}$．

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