Question

[See you again]

Cho biết S= \(\frac{1}{101}\)+ \(\frac{1}{102}\)+...+\(\frac{1}{130}\)

Chứng minh \(\frac{1}{4}\)< S < \(\frac{91}{330}\)

Answer 1

[Lâm]

S = 0,3775118592

Answer 2

[Mạnh Lê]

S = \(\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{130}\right)+\left(\frac{1}{111}+...+\frac{1}{120}\right)+\left(\frac{1}{121}+...+\frac{1}{130}\right)\)

\(\Rightarrow\frac{1}{110.10}+\frac{1}{120.10}+\frac{1}{130.10}=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}>\frac{1}{12}+\frac{2}{12}=\frac{1}{4}\)

= \(\frac{1}{11}+\frac{1}{13}>\frac{2}{12}\)

\(\Rightarrow S>\frac{1}{4}\left(1\right)\)

\(S=\left(\frac{1}{101}+\frac{1}{130}\right)+\left(\frac{1}{102}+\frac{1}{129}\right)+...+\left(\frac{1}{115}+\frac{1}{116}\right)\)( có 15 cặp )

\(=\frac{231}{101.130}+\frac{231}{102.129}+...+\frac{231}{115.116}=231\)

\(\left(\frac{1}{101.130}+\frac{1}{102.129}+...+\frac{1}{115.116}\right)\)

Ta nhận xét tích 101.130 có giá trị nhỏ nhất :

xét : 102.129 = (101+1).(130-1) = 101.129 = 101.130 - 101 + 130 - 1 = 101.130 + 28 > 101.130 

Tương tự các cặp cộng lại , ta có : \(\frac{1}{101.130}+\frac{1}{129.102}+...+\frac{1}{115.116}< \frac{1}{101.130.15}\)

\(\Rightarrow S=\frac{231.1}{101.130.15}=\frac{693}{2626}< \frac{91}{330}\)

Từ (1)(2) \(\Rightarrow\)ĐPCM