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
Những câu hỏi liên quan
chứng minh rằng với mọi n thuộc Z ta luôn \(\frac{1}{1.6}\)+ \(\frac{1}{6.11}\)+\(\frac{1}{11.16}\)+........+\(\frac{1}{\left(5n+1\right).\left(5n+6\right)}\)=\(\frac{n+1}{5n+6}\)
giúp mình đi sớm nhé
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 tỏ rằng với mọi n thuộc N ta luôn có
\(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+....+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5n+6-1}{5n+6}\)
\(=\dfrac{n+1}{5n+6}=VP\)
Đúng 2
Bình luận (0)
Giúp :3
Chứng minh rằng với mọi n \(\varepsilon\) N, ta luôn có:
\(\frac{1}{1.6}\)+ \(\frac{1}{6.11}\) +\(\frac{1}{11.16}\) +...+ \(\frac{1}{\left(5n+1\right)\left(5n+6\right)}\) = \(\frac{n+1}{5n+6}\)
B nào nhanh và đúng nhất t tick cho :33
Đặt A = \(\frac{1}{1.6}+\frac{1}{6.11}+..+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
5A = \(\frac{5}{1.6}+\frac{5}{6.11}+..+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\)
= \(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+..+\frac{1}{5n+1}-\frac{1}{5n+6}\)
= \(\frac{1}{1}-\frac{1}{5n+6}=\frac{5n+6-1}{5n+6}=\frac{5n+5}{5n+6}=\frac{5\left(n+1\right)}{5n+6}\)
=> A = \(=\frac{5\left(n+1\right)}{5n+6}:5=\frac{5\left(n+1\right)}{5n+6}\cdot\frac{1}{5}=\frac{n+1}{5n+6}\)
VẬy VT = VP ĐT Đ CM
Đúng 0
Bình luận (0)
Chứng tỏ
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+....+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)=\(\frac{n+1}{5n+6}\)
Các bn giúp mình kiểm tra với thấy ko tự tin về bài làm của mikn quá
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 )
Đúng 0
Bình luận (1)
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
Tính:
\(D=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right).\left(5n+6\right)}\)
Tính đầy đủ hộ mik vs. Mik đag cần gấp
D = \(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
= \(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}.\frac{5n+5}{5n+6}\)
= \(\frac{n+1}{5n+6}\)
Đúng 0
Bình luận (0)
Đề thi hsg nè, ai giúp làm mik làm đi.
CMR với mọi số tự nhiên n thì ta luôn có:
\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\)
Ta có:\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
\(=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left(1-\frac{1}{5n+6}\right)\)
\(=\frac{1}{5}.\left(\frac{5n+5}{5n+6}\right)=\frac{n+1}{5n+6}\left(\text{đ}pcm\right)\)
Đúng 0
Bình luận (0)
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