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\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)

\(=\dfrac{n+1}{5n+6}=VP\)

1 tháng 4 2017

\(\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)

\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)\(=\dfrac{1}{5}\cdot\left(\dfrac{5n+6}{5n+6}-\dfrac{1}{5n+6}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{5\left(n+1\right)}{5n+6}=\dfrac{n+1}{5n+6}=VP\)

1 tháng 4 2017

Thank bn nhiều nha, nhưng mà cho mk hs Vp=? vậy ạ.ngaingung

2 tháng 4 2016

tk mị̣̣̉̉̉̉̉̉̀̉̃́́́nh nhe !

19 tháng 1 2016

câu hỏi tương tự có đó bạn, bạn vào tham khảo nhe!

24 tháng 2 2017

D = \(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)

\(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)

\(\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)

\(\frac{1}{5}.\frac{5n+5}{5n+6}\)

\(\frac{n+1}{5n+6}\)

28 tháng 3 2017

Đặt :

\(A=\dfrac{3}{9.14}+\dfrac{3}{14.19}+......................+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)

\(A.\dfrac{5}{3}=\dfrac{5}{9.14}+\dfrac{5}{14.19}+..................+\dfrac{5}{\left(5n-1\right)\left(5n+1\right)}\)

\(A.\dfrac{5}{3}=\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+..................+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\)

\(A.\dfrac{5}{3}=\dfrac{1}{9}-\dfrac{1}{5n+4}\)

\(A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right):\dfrac{3}{5}\)

\(A=\left(\dfrac{1}{9}-\dfrac{1}{5n+\text{4}}\right).\dfrac{3}{5}\)

\(A=\dfrac{1}{9}.\dfrac{3}{5}-\dfrac{1}{5n+4}.\dfrac{3}{5}\)

\(A=\dfrac{1}{15}-\dfrac{1}{5.\left(5n+4\right)}\)

\(\Rightarrow A< \dfrac{1}{15}\)

\(\Rightarrowđpcm\)

Chúc bn học tốt!!!!!!!!!!

17 tháng 3 2017

kệ!! cái loại người chỉ dc cá mách lẻo là ko ai bằng! ra kia cho người khác trả lời câu hỏi!! chắn đường chắn lối tốn cả diện tích!!

17 tháng 3 2017

Ra chỗ khác ngay!!

8 tháng 8 2023

 CM:  \(\dfrac{1}{1.6}\)\(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+6}\)

A = \(\dfrac{1}{5}\)(\(\dfrac{5}{1.6}\) + \(\dfrac{5}{6.11}\)+...+ \(\dfrac{5}{\left(5n+1\right).\left(5n+6\right)}\)

A = \(\dfrac{1}{5}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{6}\)\(\dfrac{1}{6}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{5n+1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\) .( \(\dfrac{1}{1}\) - \(\dfrac{1}{5n+6}\))

A = \(\dfrac{1}{5}\)\(\dfrac{5n+6-1}{5n+6}\)

A = \(\dfrac{1}{5}\)\(\dfrac{5n+5}{5n+6}\)

A = \(\dfrac{1}{5}\) . \(\dfrac{5.\left(n+1\right)}{5n+6}\)

A = \(\dfrac{n+1}{5n+6}\)

\(\dfrac{1}{1.6}\) + \(\dfrac{1}{6.11}\)\(\dfrac{1}{11.16}\)+...+ \(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\dfrac{n+1}{5n+1}\) (đpcm)

 

 

8 tháng 8 2023

\(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)

\(A=\dfrac{1}{5}\left[1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right]\)

\(A=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)

\(A=\dfrac{1}{5}\left(\dfrac{5n+6-1}{5n+6}\right)=\dfrac{1}{5}\left(\dfrac{5n+5}{5n+6}\right)=\dfrac{1}{5}.5\left(\dfrac{n+1}{5n+6}\right)=\dfrac{n+1}{5n+6}\)

\(\Rightarrow dpcm\)