so sánh : A= 1/1.6+1/6.11+1/11.16+....+ 1/ (5n+1). (5n+6) với B= n+1/5n+6
Những câu hỏi liên quan
Chứng minh 1/1.6+1/6.11+1/11.16+...+1/(5n+1)(5n+6)=n+1/5n+6
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)
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\(A=\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(A=\dfrac{1}{5}\left[1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right]\)
\(A=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(A=\dfrac{1}{5}\left(\dfrac{5n+6-1}{5n+6}\right)=\dfrac{1}{5}\left(\dfrac{5n+5}{5n+6}\right)=\dfrac{1}{5}.5\left(\dfrac{n+1}{5n+6}\right)=\dfrac{n+1}{5n+6}\)
\(\Rightarrow dpcm\)
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\(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+....+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
C = 1/1.6+1/6.11+1/11.16+.....+1/(5n+1).(5n+6) n thuoc N
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 )
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Chứng minh rằng với mọi n thuộc N ta luôn có:
1/1.6 + 1/6.11 + 1/11.16 + ......+ 1/( 5n + 1) (5n + 6) = n+1/ 5n + 6
chứng tỏ rằng với mọi n thuộc N ta luôn có
\(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+....+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)
\(=\dfrac{n+1}{5n+6}=VP\)
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Chứng minh rằng mọi n€N ta luôn có: 1/1.6+1/6.11+1/11.16+.........+1/(5n+1)(5n+6)=n+1/5n+6
Xem chi tiết
1. Tính tổng
a, A=1/2.3 + 1/3.4 + ... + 1/99.100
b, B= 5/1.4 + 5/4.7 + ... + 5/100.103
c, C= 1/15 +1/35 + ... + 1/2499
d, D=1/1.6 + 1/6.11 + 1/11.16 + ... +1/(5n+1).(5n+6)
mn ơi mình đang cần gấp
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
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CMR: mọi n thuộc N ta có
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5v+1\right).\left(5n+6\right)}=\frac{n+1}{5n+6}\)
Chứng minh rằng với mọi n \(\in\) N ta luôn có:
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\)
Heo mi pờ lít
câu hỏi tương tự có đó bạn, bạn vào tham khảo nhe!
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