Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt S=\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)
Ta thấy S có 40 số hạng
ta có:
S=\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)=\(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)\)
\(+\left(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}\right)\)(mỗi 1 nhóm có 100 số hạng)
>\(\left(\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{70}+...+\frac{1}{70}\right)+\left(\frac{1}{80}+...+\frac{1}{80}\right)\)(mỗi 1 nhóm có 10 số hạng)
=\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\)=\(\frac{533}{840}\)>\(\frac{490}{840}\)=\(\frac{7}{12}\)
vậy S>\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)(đpcm)
Ta có :
\(\frac{7}{12}\)= \(\frac{4}{12}\)+ \(\frac{3}{12}\)= \(\frac{1}{3}\)+ \(\frac{1}{4}\)= \(\frac{20}{60}\)+ \(\frac{20}{80}\)
\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ .... + \(\frac{1}{79}\)+ \(\frac{1}{80}\)= (\(\frac{1}{41}\)+ \(\frac{1}{42}\)+ \(\frac{1}{43}\)+ ....+\(\frac{1}{60}\)) + ( \(\frac{1}{61}\)+ \(\frac{1}{62}\)+...+\(\frac{1}{79}\)+\(\frac{1}{80}\))
Do \(\frac{1}{41}\)>\(\frac{1}{42}\)>....>\(\frac{1}{60}\)
=> ( \(\frac{1}{41}\)+ \(\frac{1}{42}\)+...+\(\frac{1}{60}\)) > \(\frac{1}{60}\)+...+\(\frac{1}{60}\)= \(\frac{20}{60}\)
Vậy : \(\frac{1}{61}\)> \(\frac{1}{62}\)>....>\(\frac{1}{79}\)>\(\frac{1}{80}\)
=> ( \(\frac{1}{61}\)+\(\frac{1}{62}\)+...+\(\frac{1}{79}\)+ \(\frac{1}{80}\)) > \(\frac{1}{80}\)+...+ \(\frac{1}{80}\)= \(\frac{20}{80}\)
Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+\(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)
Vậy : \(\frac{1}{41}\)+ \(\frac{1}{42}\)+....+ \(\frac{1}{79}\)+ \(\frac{1}{80}\)> \(\frac{20}{60}\)+ \(\frac{20}{80}\)= \(\frac{7}{12}\)
=> ĐPCM
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm
1 + 1/2 + 1/3 + ... + 1/62 + 1/63 + 1/64
= 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + ... + 1/16) + (1/17 + 1/18 + ... + 1/32) + (1/33 + 1/34 + ... + 1/64)
> 1 + 1/2 + 1/4 × 2 + 1/8 × 4 + 1/16 × 8 + 1/32 × 16 + 1/64 × 32
> 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
> 1 + 1/2 × 6
> 1 + 3
> 4
1 + 1/2 + 1/3 + ... + 1/62 + 1/63 + 1/64
= 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + 1/10 + ... + 1/16) + (1/17 + 1/18 + ... + 1/32) + (1/33 + 1/34 + ... + 1/64)
> 1 + 1/2 + 1/4 × 2 + 1/8 × 4 + 1/16 × 8 + 1/32 × 16 + 1/64 × 32
> 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2
> 1 + 1/2 × 6
> 1 + 3
> 4
a) Ta có:
\(\frac{1}{n-1}-\frac{1}{n}=\frac{n-\left(n-1\right)}{n\left(n-1\right)}=\frac{1}{n\left(n-1\right)}>\frac{1}{n.n}=\frac{1}{n^2}\left(1\right)\)
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}< \frac{1}{n.n}=\frac{1}{n^2}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra:
\(\frac{1}{n\left(n-1\right)}>\frac{1}{n^2}>\frac{1}{n\left(n+1\right)}\)
Hay \(\frac{1}{n-1}-\frac{1}{n}>\frac{1}{n^2}>\frac{1}{n}-\frac{1}{n+1}\) (Đpcm)