Ta có :3²/8*11+3²/11*14+3²/14*17+...............+3²/197*200

=3*(3/8*11+3/11*14+3/14*17+..............+3/197*200)

=3*(1/8-1/11+1/11-1/14+1/14-1/17+..................+1/197-1/200)

=3*(1/8-1/200) 

=3* (3/25)

=9/25

\(=3\left(\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{197\cdot200}\right)\)

\(=3\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(=3\cdot\dfrac{192}{1600}=\dfrac{9}{25}\)

\(A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{107.111}\)

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)

\(A=\frac{1}{3}-\frac{1}{111}\)

\(A=\frac{12}{37}\)

mà dài quá bạn ơi ban tách ra thành nhiều câu hỏi đi thế này trả lời lâu lắm

b: \(27D=3^{14}+3^{17}+...+3^{2024}\)

\(\Leftrightarrow26D=3^{2024}-3^{11}\)

hay \(D=\dfrac{3^{2024}-3^{11}}{26}\)

c: \(25E=-5^4-5^6-...-5^{1002}\)

\(\Leftrightarrow24E=-5^{1002}+5^2\)

hay \(E=\dfrac{-5^{1002}+5^2}{24}\)

sai nha

\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(=\)\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}\)

\(=\frac{10}{20}-\frac{1}{20}\)

\(=\frac{9}{20}\)

Tk giúp !!

a: =35/17-18/17-9/5+4/5

=1-1=0

b: =-7/19(3/17+8/11-1)

=7/19*18/187=126/3553

c: =26/15-11/15-17/3-6/13

=1-6/13-17/3

=7/13-17/3=-200/39

a) Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{17.20}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(=\frac{1}{2}-\frac{1}{20}< \frac{1}{2}\)

Vậy A<\(\frac{1}{2}\).

b) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

                  \(\frac{1}{3^2}< \frac{1}{2.3}\)

                  \(\frac{1}{4^2}< \frac{1}{3.4}\)

                    ...

                   \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< 1-\frac{1}{100}< 1\)

Vậy \(B< 1\).