Cho A =( 30;42;56;72;90;110;132;156;182;210) B=(15;35;63;99;143;195) Chứng tỏ rằng tổng các số nghịch đảo của các phần tử thuộc tập hợp A đúng bằng tổ

Đáp án:

$B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\\
=\frac{1}{2}\left ( \frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15} \right )\\
=\frac{1}{2}\left ( \frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+\frac{13-11}{11.13}+\frac{15-13}{13.15} \right )\\
=\frac{1}{2}\left ( \frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right )\\
=\frac{1}{2}.\left ( \frac{1}{3}-\frac{1}{15} \right )\\
=\frac{1}{2}.\frac{4}{15}\\
=\frac{2}{15}\\
A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\\
=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\\
= \frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\\
=\frac{1}{5}-\frac{1}{15}\\
=\frac{2}{15}\\
\Rightarrow A=B=\frac{2}{15}$