Chung minh
1/1.6+1/11.16+............+1/(5n+1).(5n+6)=n+1/5n+6
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
PN
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
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
\(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(A=\dfrac{1}{5}\left[1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right]\)
\(A=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(A=\dfrac{1}{5}\left(\dfrac{5n+6-1}{5n+6}\right)=\dfrac{1}{5}\left(\dfrac{5n+5}{5n+6}\right)=\dfrac{1}{5}.5\left(\dfrac{n+1}{5n+6}\right)=\dfrac{n+1}{5n+6}\)
\(\Rightarrow dpcm\)
N
1
9 tháng 8 2019
Ta có : \(A=\frac{1}{1\cdot6}+\frac{1}{6\cdot11}+\frac{1}{11\cdot16}+...+\frac{1}{(5n+1)(5n+6)}\)
\(=\frac{1}{5}\cdot\left[\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{(5n+1)(5n+6)}\right]\)
\(=\frac{1}{5}\cdot\left[1-\frac{1}{5n+6}\right]=\frac{1}{5}\cdot\frac{5n+6-1}{5n+6}=\frac{1}{5}\cdot\frac{5(n+1)}{5n+6}=\frac{n+1}{5n+6}\)
NT
1
AK
10 tháng 3 2018
C = 1/1 . 6 + 1/6 . 11 + 1/11 . 16 + ...+ 1/( 5n + 1 ) . ( 5n + 6 )
C = 1/5 . ( 5/1 . 6 + 5/6 . 11 + 5/11 . 16 + ...+ 5/( 5n + 1 ) . ( 5n + 6 ) )
C = 1/5 . ( 1 - 1/6 + 1/6 - 1/11 + 1/11 - 1/16 + ...+ 1/5n + 1 - 1/5n + 6 )
C = 1/5 . ( 1 - 1/5n + 6 )
C = 1/5 . 1 - 1/5 . 1/5n + 6
C = 1/5 - 1/ 5 . ( 5n + 6 )
1 tháng 3 2022
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)
\(=\dfrac{n+1}{5n+6}=VP\)
Dãy số viết theo quy luật mà bạn, nhiều bài thế này rồi, bạn muốn cụ thể thì vào Google mà tìm ...