Ta có : 5¹ + 5² +...+ 5²⁰¹⁶ 

= (5+5²+5³)+(5⁴+5⁵+5⁶)+...+(5²⁰¹⁴+5²⁰¹⁵+5²⁰¹⁶)

= 5(1+5+5²)+5³(1+5+5²)+...+5²⁰¹³(1+5+5²)

= 5. 31 + 5³.31 +...+5²⁰¹³.31

= 31(5+5²+...+5²⁰¹³) chia hết cho 31

Vây 5¹ + 5² + ...+ 5²⁰¹⁶ chia hết cho 31

Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được

tôi biết câu này nè

LÀM TẮT NHÉ :

\(C=\frac{3}{4}\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\right)\))

\(D=\frac{4}{5}\left(\frac{1}{2}-\frac{1}{7}+...+\frac{1}{102}-\frac{1}{107}\right)\)

tương tự với các phần còn lại

Cho S=\(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{22.27}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{22}-\frac{1}{27}\)

\(=\frac{1}{2}-\frac{1}{27}=\frac{25}{54}\)

kb nha mn!

\(\frac{5}{2}-\frac{5}{7}+\frac{5}{7}-\frac{5}{12}+\frac{5}{12}-\frac{5}{17}+\frac{5}{17}-\frac{5}{22}+\frac{5}{22}-\frac{5}{29}=\frac{5}{2}-0-0-0-0-\frac{5}{29}=\frac{5}{2}-\frac{5}{29}=\frac{145}{58}-\frac{10}{58}=\frac{135}{58}\)

TA CÓ:   \(\frac{1}{n}-\frac{1}{n+5}=\frac{n+5-n}{n\left(n+5\right)}=\frac{5}{n\left(n+5\right)}\)

Thay vào biểu thức trên , ta được:

\(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{907.1002}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{907}-\frac{1}{1002}\)

\(=\frac{1}{2}-\frac{1}{1002}=\frac{250}{501}\)

\(N=2015+\frac{10}{2.7}+\frac{10}{7.12}+\frac{10}{12.17}+\frac{10}{17.22}\)

     \(=2\left(1007,5+\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}\right)\)

     \(=2\left(1007,5+\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+\frac{22-17}{17.22}\right)\)

     \(=2\left(1007,5+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}\right)\)

     \(=2\left(1007,5+\frac{1}{2}-\frac{1}{22}\right)\)

     \(=2015+1-\frac{1}{11}\)

     \(=\frac{22175}{11}\)

N = \(2015+\frac{10}{2,7}+\frac{10}{7,12}+\frac{10}{12,17}+\frac{10}{17,22}=2021.510611\)

\(N=2015+\frac{10}{2\cdot7}+\frac{10}{7\cdot12}+\frac{10}{12\cdot17}+\frac{10}{17\cdot22}=2015+2\left(\frac{5}{2\cdot7}+\frac{5}{7\cdot12}+\frac{5}{12\cdot17}+\frac{5}{17\cdot22}\right)\)\(=2015+2\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}\right)\)

\(=2015+2\left[\left(\frac{1}{2}-\frac{1}{22}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{17}-\frac{1}{17}\right)\right]\)

\(=2015+2\left[\left(\frac{11}{22}-\frac{1}{22}\right)+0+...+0\right]=2015+2\cdot\frac{10}{22}=2015+\frac{10}{11}\)

xong rồi đó, cái còn lại thì bạn quy đồng rồi cộng nha

mk cũng ko chắc do tính nhẩm

Đặt
\(A=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+\frac{5}{17.22}+\frac{5}{22.27}+\frac{5}{27.32}+\frac{5}{32.37}\)

\(A=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{22}+\frac{1}{22}-\frac{1}{27}+\frac{1}{27}-\frac{1}{32}+\frac{1}{32}-\frac{1}{37}\)

\(A=\frac{1}{2}-\frac{1}{37}=\frac{35}{74}\)

\(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{32}-\frac{1}{37}\)

\(=\frac{1}{2}-\frac{1}{37}\)

\(=\frac{37}{74}-\frac{2}{74}=\frac{35}{74}\)

\(=\frac{1}{2}.\frac{1}{7}+\frac{1}{7}.\frac{1}{12}...+\frac{1}{32}.\frac{1}{37}\)

= \(\frac{1}{2}+\frac{1}{37}\)

= \(\frac{37}{74}+\frac{2}{74}\)

\(\frac{39}{74}\)

tíck mik nha, mik nhanh nhất

\(\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+....+\frac{1}{2012.2017}\)

\(=\frac{1}{5}\left(\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+....+\frac{5}{2012.2017}\right)\)

\(=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+....+\frac{1}{2012}-\frac{1}{2017}\right)\)

\(=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{2017}\right)\)

\(=\frac{1}{5}.\frac{2015}{4034}=\frac{403}{4034}\)

ĐẶT A=DÃY SỐ TRÊN=>5A=5/2.7+........+5/2012.2017

=>A=1/2-1/7........-1/2012-1/2017 RÚT GỌN TA ĐƯỢC A=1/2-1/2017

Đặt \(M=\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+...+\frac{1}{2012.2017}\)

\(\Rightarrow\) \(5M=5.(\frac{1}{2.7}+\frac{1}{7.12}+\frac{1}{12.17}+...+\frac{1}{2012.2017})\)

           \(=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{2012.2017}\)

           \(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{2012}-\frac{1}{2017}\)

           \(=\frac{1}{2}-\frac{1}{2017}\)

\(\Rightarrow\) \(M=(\frac{1}{2}-\frac{1}{2017}):5\)

         \(=\frac{1}{10}-\frac{1}{10085} \)

Vậy \(M=\frac{1}{10}-\frac{1}{10085}\)

\(B=\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{87.92}\)

\(\Leftrightarrow\frac{5}{3}B=\frac{5}{2.7}+\frac{5}{7.12}+\frac{5}{12.17}+...+\frac{5}{87.92}\)

\(\Leftrightarrow\frac{5}{3}B=\frac{7-2}{2.7}+\frac{12-7}{7.12}+\frac{17-12}{12.17}+...+\frac{92-87}{87.92}\)

\(\Leftrightarrow\frac{5}{3}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{87}-\frac{1}{92}\)

\(\Leftrightarrow\frac{5}{3}B=\frac{1}{2}-\frac{1}{92}\)

\(\Leftrightarrow\frac{5}{3}B=\frac{46}{92}-\frac{1}{92}\)

\(\Leftrightarrow\frac{5}{3}B=\frac{45}{92}\)

\(\Leftrightarrow B=\frac{45}{92}\times\frac{3}{5}\)

\(\Leftrightarrow B=\frac{27}{92}\)

CTV làm màu vc :)

\(B=\frac{3}{2.7}+\frac{3}{7.12}+\frac{3}{12.17}+...+\frac{3}{87.92}\)

\(B=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{87}-\frac{1}{92}\right)\)

\(B=\frac{3}{5}.\left(\frac{1}{2}-\frac{1}{92}\right)\)

\(B=\frac{3}{5}.\frac{45}{92}\)

\(B=\frac{27}{92}\)

\(\Rightarrow C=\frac{10}{5}\left(\frac{1}{7.12}+\frac{1}{12.17}+\frac{1}{17.22}+...+\frac{1}{502.507}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+....+\frac{1}{507}-\frac{1}{507}\right)\)

\(\Rightarrow C=2.\left(\frac{1}{7}-\frac{1}{507}\right)=2.\frac{1}{7}-2.\frac{1}{507}=\frac{2}{7}-\frac{2}{507}\)