Tổng S có 30 số hạng.

Ta có: \(\frac{5}{49}< \frac{5}{48}< \frac{5}{47}< ...< \frac{5}{22}< \frac{5}{21}< \frac{5}{20}.\)

Suy ra : \(3=30\cdot\frac{5}{50}< 30\cdot\frac{5}{49}< S< 30\cdot\frac{5}{20}=\frac{15}{5}< 8\)

Vậy 3<S<8. ĐPCM.

\(S=\frac{5}{20}+\frac{5}{21}+..........+\frac{5}{49}\)

\(=5\left(\frac{1}{20}+\frac{1}{21}+.......+\frac{1}{49}\right)\)

Mà \(\frac{1}{20}>\frac{1}{49};\frac{1}{21}>\frac{1}{49};.........;\frac{1}{49}=\frac{1}{49}\)

\(\Leftrightarrow5\left(\frac{1}{20}+\frac{1}{21}+.....+\frac{1}{49}\right)>5\left(\frac{1}{49}+\frac{1}{49}+.......+\frac{1}{49}\right)\)

\(\Leftrightarrow S>5.\frac{30}{49}\)

\(\Leftrightarrow S>3\frac{3}{49}\)

\(\Leftrightarrow S>3\left(1\right)\)

Lại có :

\(\frac{1}{20}=\frac{1}{20};\frac{1}{21}< \frac{1}{20};.......;\frac{1}{49}< \frac{1}{20}\)

\(\Leftrightarrow S=5\left(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{49}\right)< 5\left(\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}\right)\)

\(\Leftrightarrow S< 5.\frac{30}{20}=7\frac{1}{2}\)

\(\Leftrightarrow S< 8\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow3< S< 8\)

https://olm.vn/hoi-dap/tim-kiem?q=+++++++++++Cho+s=5/20+5/21+5/22+5/23+...+5/49.ch%E1%BB%A9ng+minh+r%E1%BA%B1ng:+3+%3C+s+%3C+8&id=376641

tham khảo nhé bn

https://olm.vn/hoi-dap/tim-kiem?q=Cho+S+=5/20+5/21+5/22+5/23+..........+5/49.+Ch%E1%BB%A9ng+minh+r%E1%BA%B1ng+3%3CS%3C8&id=68660

tham khảo nhé bn

\(S=5.\left(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{49}\right)\)

Xét \(A=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{49}\). Chứng minh 3/5 < A < 8/5

+ Có: \(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}\frac{3}{5}\Rightarrow S>3\)  (2)

Từ (1)(2) => 3 < S < 8

Này Trần Thị Loan à, tớ thấy cậu nên

thay chữ "xét" ở chỗ "xét A" thành chữ"đặt"

nghe hợp lý hơn.

Tách từng nhóm 2 số ra mà làm