tính các tổng sau một cách hợp lý : B= 1/1.6 + 1/6.11 + 1/11.16 + .........+ 1/(5n+1) . (5n+6) giúp mình nhé mình gấp lắm rồi làm ơn
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17 tháng 6 2023
a: =1/2-1/3+1/3-1/4+...+1/99-1/100
=1/2-1/100=49/100
b; =5/3(1-1/4+1/4-1/7+...+1/100-1/103)
=5/3*102/103
=510/309=170/103
c: =1/2(1/3-1/5+1/5-1/7+...+1/49-1/51)
=1/2*16/51=8/51
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)
ND
Nguyễn Đức Trí
VIP
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 )
S
24 tháng 2 2017
D = \(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
= \(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}.\frac{5n+5}{5n+6}\)
= \(\frac{n+1}{5n+6}\)
IM
30 tháng 8 2016
Ta có
\(\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left[\frac{\left(5n+6\right)-1}{\left(5n+6\right)}\right]\)
\(=\frac{1}{5}.\frac{5n+5}{5n+6}\)
\(=\frac{n+1}{5n+6}\)
\(\Rightarrow\frac{1}{1.6}+\frac{1}{6.11}+......+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\) ( đpcm )
\(B=\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{5n+1}+\frac{1}{5n+6}\)
\(B=\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{5n+1}-\frac{1}{5n+6}\)
\(B=\frac{1}{1}-\frac{1}{5n+6}=\frac{5n+5}{5n+6}\)