S= 1/2^2 + 1/3^3 + 1/4^4 +….+ 1/99^2 . chứng minh 49/100 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " S= 1/2^2 + 1/3^3 + 1/4^4 +....+ 1/99^2 . chứng minh 49/100
S= 1/2×2 + 1/3×3 + 1/4×4+….+1/99×99
+) ta có : S< 1/1×2 + 1/2×3+1/3×4 +….+ /98×99
S<1-1/2+1/2-1/3+1/3-1/4+…..+1/98-1/99
S<1-1/99
S< 98/99 (1)
+) ta lại có: S> 1/2×3+1/3×4+1/4×5 +….+1/99×100
S>1/2-1/3+1/3-1/4+…+1/99-1/100
S>1/2 -1/100
S>49/100 (2)
từ (1) và (2) => 49/100< S < 98/99 .
Tham khảo
Xét `S`
`⇒\frac{1}{2.3}+\frac{1}{3.4}+…+\frac{1}{99.100}<S<\frac{1}{1.2}+\frac{1}{2.3}+…+\frac{1}{98.99}`
Áp dụng `\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+2}`
`⇒\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{99}-\frac{1}{100}<S<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+…+\frac{1}{98}-\frac{1}{99}`
`⇒\frac{1}{2}-\frac{1}{100}<S<1-\frac{1}{99}`
`⇒\frac{49}{100}<S<\frac{98}{99}`
`\text{©CBT}`