每日一题[724]数列求和

已知正项数列$\{a_n\}$满足$a_1=\dfrac 32$,$a_{n+1}^2-a_n^2=\dfrac{1}{(n+2)^2}-\dfrac{1}{n^2}$,记数列$\{a_n\}$的前$n$项和为$S_n$,则$\dfrac{1}{S_1}-\dfrac{1}{S_3}+\dfrac{1}{S_5}-\cdots -\dfrac{1}{S_{2007}}+\dfrac{1}{S_{2009}}$的值为______.


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分析与解 $\dfrac{1006}{2011}$.

由累加法知$$a_n^2=\dfrac 1{(n+1)^2}+\dfrac 1{n^2}+1,$$所以$a_n=1+\dfrac 1n-\dfrac{1}{n+1}$.直接求和得$S_n=n+1-\dfrac 1{n+1}$,所以$$\dfrac 1{S_n}=\dfrac {n+1}{n(n+2)}=\dfrac 12\left(\dfrac 1n+\dfrac 1{n+2}\right).$$所求的式子记为$M$,有$$2M=1+\dfrac 13-\left(\dfrac 13+\dfrac 15\right)+\left(\dfrac 15+\dfrac 17\right)-\left(\dfrac 17+\dfrac 19\right)+\cdots+\left(\dfrac 1{2009}+\dfrac 1{2011}\right)=\dfrac{2012}{2011},$$所以$M=\dfrac{1006}{2011}$.

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