cho S = 1/2^2+1/3^2+1/4^2+…+1/9^2 cmr 2/5 05/07/2021 Bởi Nevaeh cho S = 1/2^2+1/3^2+1/4^2+…+1/9^2 cmr 2/5 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho S = 1/2^2+1/3^2+1/4^2+...+1/9^2 cmr 2/5
Ta có S = 1/2² + 1/3² + 1/4² +…..+ 1/9² < 1/2² + 1/2 × 3 + 1/3 × 4 +…..+ 1/8 × 9 = 1/2² + 1/2 – 1/3 + 1/3 – 1/4+…..+1/8 – 1/9 = 1/4 + 1/2 – 1/9 = 23/36 < 32/36 = 8/9 (1) Ta lại có S = 1/2² + 1/3² + 1/4² +…..+ 1/9² > 1/2² + 1/3 × 4 + 1/4 × 5+……+ 1/9 × 10 = 1/2² + 1/3 – 1/4 + 1/4 – 1/5 +……+ 1/9 – 1/10 = 1/2² + 1/3 – 1/10 = 19/20 > 8/20 = 2/5 (2) Từ (1)(2) cho ta đpcm. Bình luận
$S= \frac{1}{2²}+\frac{1}{3²}+…+\frac{1}{9²}$ Ta có: $S> \frac{1}{2.3}+\frac{1}{3.4}+…+\frac{1}{9.10}$ ⇔ $S> \frac{1}{2}-\frac{1}{10}= \frac{2}{5}$ (*) Ta có: $S< \frac{1}{1.2}+\frac{1}{2.3}+…+\frac{1}{8.9}$ ⇔ $S< 1-\frac{1}{9}=\frac{8}{9}$ (**) Từ (*) và (**) ⇒ $\frac{2}{5}< S< \frac{8}{9}$ Bình luận
Ta có S = 1/2² + 1/3² + 1/4² +…..+ 1/9²
< 1/2² + 1/2 × 3 + 1/3 × 4 +…..+ 1/8 × 9
= 1/2² + 1/2 – 1/3 + 1/3 – 1/4+…..+1/8 – 1/9
= 1/4 + 1/2 – 1/9
= 23/36 < 32/36 = 8/9 (1)
Ta lại có S = 1/2² + 1/3² + 1/4² +…..+ 1/9²
> 1/2² + 1/3 × 4 + 1/4 × 5+……+ 1/9 × 10
= 1/2² + 1/3 – 1/4 + 1/4 – 1/5 +……+ 1/9 – 1/10
= 1/2² + 1/3 – 1/10
= 19/20 > 8/20 = 2/5 (2)
Từ (1)(2) cho ta đpcm.
$S= \frac{1}{2²}+\frac{1}{3²}+…+\frac{1}{9²}$
Ta có: $S> \frac{1}{2.3}+\frac{1}{3.4}+…+\frac{1}{9.10}$
⇔ $S> \frac{1}{2}-\frac{1}{10}= \frac{2}{5}$ (*)
Ta có: $S< \frac{1}{1.2}+\frac{1}{2.3}+…+\frac{1}{8.9}$
⇔ $S< 1-\frac{1}{9}=\frac{8}{9}$ (**)
Từ (*) và (**) ⇒ $\frac{2}{5}< S< \frac{8}{9}$