Question

15．已知數(shù)列{a_n}為公差不為0的等差數(shù)列，S_n為前n項(xiàng)和，a₅和a₇的等差中項(xiàng)為11，且a₂•a₅=a₁•a₁₄．令b_n=$\frac{1}{{{a_n}•{a_{n+1}}}}$，數(shù)列{b_n}的前n項(xiàng)和為T_n．求a_n及T_n．

Answer 1

分析 設(shè)等差數(shù)列{a_n}的公差為d，依題意，可求得數(shù)列{a_n}的首項(xiàng)與公差，從而可得其通項(xiàng)a_n；再由b_n=$\frac{1}{{{a_n}•{a_{n+1}}}}$，利用裂項(xiàng)法可求得數(shù)列{b_n}的前n項(xiàng)和為T_n．

解答 解：因?yàn)閧a_n}為等差數(shù)列，設(shè)公差為d，則由題意得$\left\{\begin{array}{l}{a_5}+{a_7}=22⇒2{a_1}+10d=22\\{a_2}•{a_5}={a_1}•{a_{14}}⇒（{a_1}+d）（{a_1}+4d）={a_1}（{a_1}+13d）\end{array}\right.$，
整理得$\left\{\begin{array}{l}{a_1}+5d=11\\ d=2{a_1}\end{array}\right.⇒\left\{\begin{array}{l}d=2\\{a_1}=1\end{array}\right.$，
所以a_n=1+（n-1）×2=2n-1，
又${b_n}=\frac{1}{{{a_n}•{a_{n+1}}}}=\frac{1}{（2n-1）（2n+1）}=\frac{1}{2}（\frac{1}{2n-1}-\frac{1}{2n+1}）$，
所以${T_n}=\frac{1}{2}（1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+…+\frac{1}{2n-1}-\frac{1}{2n+1}）=\frac{n}{2n+1}$．

點(diǎn)評(píng) 本題考查數(shù)列的求和，著重考查解方程組求通項(xiàng)與裂項(xiàng)法求和的應(yīng)用，屬于中檔題．