Cho B=1/19^2+1/20^2+1/21^2+…..+1/380^2 Chứng minh 1/20 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " Cho B=1/19^2+1/20^2+1/21^2+.....+1/380^2 Chứng minh 1/20
Cho B=1/19^2+1/20^2+1/21^2+…..+1/380^2 Chứng minh 1/20
By Nevaeh
By Nevaeh
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 $1,2⇒\dfrac{1}{20}<B<\dfrac{1}{18}$
Vậy đpcm
Giải thích các bước giải:
Xét:
`1/19.20<1/19^2<1/18.19`
`1/20.21<1/20^2<1/19.20`
`………………………………………….`
`1/380.381<1/380^2<1/379.380`
`=>1/19.20+1/20.21+…+1/380.381<1/19^2+1/20^2+…+1/380^2<1/18.19+1/19.20+…+1/379.380`
`=>1/19-1/20+1/20-1/21+…+1/380-1/381<B<1/18-1/19+1/19-1/20+…+1/279-1/380`
`=>1/19-1/381<B<1/18-1/380`
`=>1/20<B<1/18`