Question

4．若等差數(shù)列{a_n}的前n項(xiàng)和為S_n，a₄=4，S₄=10，則數(shù)列$\left\{{\frac{1}{{\;{a_n}{a_{n+1}}\;}}}\right\}$的前2018項(xiàng)的和為$\frac{2018}{2019}$．

Answer 1

分析 由已知列式求出等差數(shù)列的首項(xiàng)和公差，得到等差數(shù)列的通項(xiàng)公式，代入$\left\{{\frac{1}{{\;{a_n}{a_{n+1}}\;}}}\right\}$，利用裂項(xiàng)相消法求和．

解答 解：由a₄=4，S₄=10，得$\left\{\begin{array}{l}{a_4}={a_1}+3d=4\\{S_4}=4{a_1}+\frac{4×3}{2}d=10\end{array}\right.$，解得$\left\{\begin{array}{l}{a_1}=1\\ d=1\end{array}\right.$，
∴a_n=n，
∴$\frac{1}{{{a_n}{a_{n+1}}}}=\frac{1}{{n（{n+1}）}}=\frac{1}{n}-\frac{1}{n+1}$，
則數(shù)列$\left\{{\frac{1}{{\;{a_n}{a_{n+1}}\;}}}\right\}$的前2018項(xiàng)的和為$（{1-\frac{1}{2}}）+（{\frac{1}{2}-\frac{1}{3}}）+（{\frac{1}{3}-\frac{1}{4}}）+…+（{\frac{1}{2017}-\frac{1}{2018}}）+（{\frac{1}{2018}-\frac{1}{2019}}）=1-\frac{1}{2019}=\frac{2018}{2019}$．
故答案為：$\frac{2018}{2019}$．

點(diǎn)評(píng) 本題考查裂項(xiàng)相消法求數(shù)列的前n項(xiàng)和，考查等差數(shù)列通項(xiàng)公式的求法，是中檔題．