Question

Дана арифметическая прогрессия a1,a2,…,a443, у которой a1=2,a443=17. найдите сумму 1/а1a2 +1/a2a3+1/a3a4+…+1/a442a443 .

Answer 1

ответ: 13

Пошаговое объяснение:

$a_n=a_1+(n-1)d$ — n-ый член арифметической прогрессии$a_{443}=a_1+442d~~\Rightarrow~~ d=\frac{a_{443}-a_1}{442}=\frac{17-2}{442}=\frac{15}{442}$

$\frac{1}{a_1a_2}+\frac{1}{a_2a_3}+\frac{1}{a_3a_4}+...+\frac{1}{a_{442}a_{443}}=\frac{1}{a_1(a_1+d)}+\frac{1}{(a_1+d)(a_1+2d)}+\frac{1}{(a_1+2d)(a_1+3d)}\\ \\+...+\frac{1}{(a_1+441d)(a_1+442d)}=-\frac{1}{d}{\bigg(\frac{a_1-(a_1+d)}{a_1(a_1+d)}+\frac{a_1+d-(a_1+2d)}{(a_1+d)(a_1+2d)}+\frac{a_1+2d-(a_1+3d)}{(a_1+2d)(a_1+3d)}$

$+...+\frac{a_1+441d-(a_1+442d)}{(a_1+441d)(a_1+442d)}\bigg)=-\frac{1}{d}\bigg(-\frac{1}{a_1}+\frac{1}{a_1+d}+\frac{1}{a_1+2d}-\frac{1}{a_1+d}+\frac{1}{a_1+3d}-\\ \\ -\frac{1}{a_1+2d}+...+\frac{1}{a_1+441d}-\frac{1}{a_1+442d}\bigg)=\frac{1}{d}\bigg(\frac{1}{a_1}-\frac{a_1}{a_1+442d}\bigg)=\frac{1}{d}\bigg(\frac{1}{a_1}-\frac{1}{a_{443}}\bigg)=\\ \\ =\frac{442}{15}\bigg(\frac{1}{2}-\frac{1}{17}\bigg)=\frac{221}{15}-\frac{26}{15}=\frac{195}{15}=13$