宝鸡中学2017届高三总复习同步微专题27:先放缩后裂项

裂项相消法是数列求和的一种重要的方法,有时候需要先进行适当的放缩,再使用裂项相消法求和。以1道宝中高三同步复习题举例说明。

1.正项数列\left\{ a_n \right\}的前n项和S_n满足:S_n^2 - (n^2+n-1)S_n - (n^2+n) = 0

(1) 求数列\left\{ a_n \right\}的通项公式a_n

(2) 令b_n = \dfrac{n+1}{(n+2)^2a_n^2},数列\left\{ b_n \right\}的前n项和为T_n,证明:对于任意的n \in N_{+},都有T_n < \dfrac{5}{64}

分析与解    (1) 由S_n^2 - (n^2+n-1)S_n - (n^2+n) = 0因式分解得:

\left[ S_n - (n^2+n) \right] \left( S_n + 1 \right) = 0

\therefore S_n = n^2+n。当n \ge 2时,a_n = S_n - S_{n-1} = 2n,且a_1 = S_1 = 2满足上式,故a_n = 2n

(2) 代入a_nb_n = \dfrac{n+1}{4(n+2)^2n^2} = \dfrac{1}{16} \left[ \dfrac{1}{n^2} - \dfrac{1}{(n+2)^2} \right]

\therefore T_n = \dfrac{1}{16} \times \left[ 1 - \dfrac{1}{3^2}  + \dfrac{1}{2^2} - \dfrac{1}{4^2} + \cdots + \dfrac{1}{n^2} - \dfrac{1}{(n+2)^2}\right] < \dfrac{1}{16} \times \dfrac54 = \dfrac{5}{64}


2.正项数列\left\{ a_n \right\}的前n项和为S_n,且S_n^2 - (n^2+n-3)S_n - 3(n^2+n) = 0

求证:\dfrac{1}{a_1(a_1+1)} + \dfrac{1}{a_2(a_2+1)} + \cdots + \dfrac{1}{a_n(a_n+1)} <\dfrac13

分析与解    S_n^2 - (n^2+n-3)S_n - 3(n^2+n) = 0,即\left[ S_n - (n^2+n) \right] \left[ S_n +3 \right] =0\therefore S_n = n^2+na_n  = 2n

考虑\dfrac{1}{a_n(a_n+1)} = \dfrac{1}{2n(2n+1)},如果直接裂项成\dfrac{1}{2n} - \dfrac{1}{2n+1},则起不到前后项相消的目的。故尝试将其先进行放缩。

注意到2n(2n+1) > (2n-1)(2n+1),故\dfrac{1}{2n(2n+1)} < \dfrac{1}{(2n-1)(2n+1)} = \dfrac12 \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right)

\therefore \dfrac{1}{a_1(a_1+1)} + \dfrac{1}{a_2(a_2+1)} + \cdots + \dfrac{1}{a_n(a_n+1)}

< \dfrac16 + \dfrac12 \left( \dfrac13 - \dfrac15 + \dfrac15 - \dfrac17 + \cdots + \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right)

< \dfrac16 + \dfrac16 = \dfrac13

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