A= 3/11*16+3/16*21+3/21*26+.....+3/61*66

\(=\frac{3}{5}\left(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\right)\)

\(=\frac{3}{5}\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\right)\)

\(=\frac{3}{5}\left(\frac{1}{11}-\frac{1}{66}\right)\)

\(=\frac{3}{5}\cdot\frac{5}{66}\)

\(=\frac{1}{22}\)

\(A=\frac{3}{11.16}+\frac{3}{16.21}+\frac{3}{21.26}+...+\frac{3}{61.66}\)

\(\Rightarrow A=\frac{3}{5}\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\right)\)

\(\Rightarrow A=\frac{3}{5}.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\right)\)

\(\Rightarrow A=\frac{3}{5}.\left(\frac{1}{11}-\frac{1}{66}\right)\)

\(\Rightarrow A=\frac{3}{5}.\frac{5}{66}\)

\(\Rightarrow A=\frac{1}{22}\)

Vậy \(A=\frac{1}{22}\)

a)   \(D=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{512}+\frac{1}{1024}\)

=>  \(2D=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...++\frac{1}{256}+\frac{1}{512}\)

=>  \(2D-D=\left(1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)

=>  \(D=1-\frac{1}{1024}\)

b)  \(Đ=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}=\frac{19}{20}\)

a) D=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\dots+\frac{1}{512}+\frac{1}{1024}.\) 

\(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\dots+\frac{1}{512}-\frac{1}{1024}\)

\(D=1-\frac{1}{1024}\)

\(D=\frac{1023}{1024}\)

\(Đ=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{18\cdot19}+\frac{1}{19\cdot20}\)

\(Đ=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\)

\(Đ=1-\frac{1}{20}\) 

\(Đ=\frac{19}{20}\)

Phần c như kiểu sai đề chỗ cuối hay sao ấy.

a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)

b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)

\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)

\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)

\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

nêu viết đề bài như thế thì nghĩ như là\(\frac{2\cdot4}{1}+\frac{2\cdot7}{4}...\)
=> hai bạn kia sai

Bạn dùng công thức này mà làm nhé :\(\frac{n}{a\times\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)

Ví dụ :\(\frac{2}{3\times5}=\frac{1}{3}-\frac{1}{5};\frac{3}{4\times7}=\frac{1}{4}-\frac{1}{7};\frac{5}{6\times11}=\frac{1}{6}-\frac{1}{11}\)

5B = 1/11 - 1/16 + 1/16 - 1/21 + 1/21 - 1/26 +.........+ 1/56 - 1/61 +1/61 - 1/66 
5B = 1/11 - 1/66 
5B = 5/66 
B = 1/66

1/66

nha bn

\(B=\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{61.66}\)

\(\Rightarrow5B=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(\Rightarrow5B=\frac{6}{66}-\frac{1}{66}=\frac{5}{66}\)

\(\Rightarrow B=\frac{1}{66}\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b)\(\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{61.66}\)

\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}\)

\(=\frac{5}{66}\)

a,\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

ta có:

\(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)

...

\(\frac{1}{99.100}=\frac{1}{99}-\frac{1}{100}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}=\frac{99}{100}\)

b,

\(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.16}+...+\frac{5}{61.66}\)

ta có:

\(\frac{5}{11.16}=\frac{16-11}{11.16}=\frac{16}{11.16}-\frac{11}{11.16}=\frac{1}{11}-\frac{1}{16}\)

\(\frac{5}{16.21}=\frac{21-16}{16.21}=\frac{21}{16.21}-\frac{16}{16.21}=\frac{1}{16}-\frac{1}{21}\)

...

\(\frac{5}{61.66}=\frac{66-61}{61.66}=\frac{66}{61.66}-\frac{61}{61.66}=\frac{1}{61}-\frac{1}{66}\)

= \(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

=\(\frac{1}{11}-\frac{1}{66}\)=\(\frac{5}{66}\)

\(=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}=\frac{6}{66}-\frac{1}{66}=\frac{5}{66}\)

Thank you

\(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}=5\left(\frac{1}{11.16}+\frac{1}{16.21}+\frac{1}{21.26}+...+\frac{1}{61.66}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+....+\frac{1}{61}-\frac{1}{66}\right)=5\left(\frac{1}{11}-\frac{1}{66}\right)=5.\frac{5}{66}=\frac{25}{66}\)

\(=\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\)

\(=\dfrac{1}{11}-\dfrac{1}{66}=\dfrac{5}{66}\)