Đáp án:
$\text{Chúc bạn học tốt}$
Giải thích các bước giải:
Ta có:$B=\dfrac{1}{19^2}+\dfrac{1}{20^2}+..+\dfrac{1}{380^2}$
Xét $B:$
$⇒B<\dfrac{1}{18×19}+\dfrac{1}{19×20}+...+\dfrac{1}{379}×\dfrac{1}{380}$
$ADCT:\dfrac{k}{n(n+k)}=\dfrac{1}{n}-\dfrac{1}{n+k}$
$⇒B<\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}+..+\dfrac{1}{379}-\dfrac{1}{380}$
$⇒B<\dfrac{1}{18}-\dfrac{1}{380}$
Vì $\dfrac{1}{18}-\dfrac{1}{380}<\dfrac{1}{18}$
$⇒B<\dfrac{1}{18}(1)$
Xét $B:$
$⇒B>\dfrac{1}{19×20}+\dfrac{1}{20×21}+...+\dfrac{1}{380×381}$
$ADCT:\dfrac{k}{n(n+k)}=\dfrac{1}{n}-\dfrac{1}{n+k}$
$⇒B>\dfrac{1}{19}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{21}+..+\dfrac{1}{380}-\dfrac{1}{381}$
$⇒B>\dfrac{1}{19}-\dfrac{1}{381}$
Vì $\dfrac{1}{19}-\dfrac{1}{381}>\dfrac{1}{19}-\dfrac{1}{380}=\dfrac{1}{20}$
$⇒B>\dfrac{1}{20} (2)$
Từ $1,2⇒\dfrac{1}{20}<B<\dfrac{1}{18}$
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