Question

7．计算：$\frac{4}{1×3}$-$\frac{8}{3×5}$+$\frac{12}{5×7}$-$\frac{16}{7×9}$+…+（-1）ⁿ⁺¹$\frac{4n}{（2n-1）（2n+1）}$=$\left\{\begin{array}{l}1+\frac{1}{2n+1}，n为奇数\\ 1-\frac{1}{2n+1}，n为偶数\end{array}\right.$．

Answer 1

分析 通过裂项法，利用n为奇数与偶数，分别求解数列的和即可．

解答 解：因为$\frac{4n}{（2n-1）（2n+1）}$=$\frac{1}{2n-1}+\frac{1}{2n+1}$，
所以当n为奇数时，
$\frac{4}{1×3}$-$\frac{8}{3×5}$+$\frac{12}{5×7}$-$\frac{16}{7×9}$+…+（-1）ⁿ⁺¹$\frac{4n}{（2n-1）（2n+1）}$
=$1+\frac{1}{3}$$-\frac{1}{3}-\frac{1}{5}$$+\frac{1}{5}+\frac{1}{7}$-…$-\frac{1}{2n-3}-\frac{1}{2n-1}$$+\frac{1}{2n-1}+\frac{1}{2n+1}$
=1+$\frac{1}{2n+1}$．
当n为偶数时，
$\frac{4}{1×3}$-$\frac{8}{3×5}$+$\frac{12}{5×7}$-$\frac{16}{7×9}$+…+（-1）ⁿ⁺¹$\frac{4n}{（2n-1）（2n+1）}$
=$1+\frac{1}{3}$$-\frac{1}{3}-\frac{1}{5}$$+\frac{1}{5}+\frac{1}{7}$-…$+\frac{1}{2n-3}+\frac{1}{2n-1}$$-\frac{1}{2n-1}-\frac{1}{2n+1}$
=1-$\frac{1}{2n+1}$．
故答案为：$\left\{\begin{array}{l}1+\frac{1}{2n+1}，n为奇数\\ 1-\frac{1}{2n+1}，n为偶数\end{array}\right.$．

点评 本题考查数列求和，裂项法的应用，考查分类讨论的应用．