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$a_{443}=a_1+442d~~Rightarrow~~ d=frac{a_{443}-a_1}{442}=frac{17-2}{442}=frac{15}{442}$

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

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