Đề thi hsg nè, ai giúp làm mik làm đi.
CMR với mọi số tự nhiên n thì ta luôn có:
\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\)
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!
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
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+....+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)=\(\frac{n+1}{5n+6}\)
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 )
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
cmr với mọi n thuộc N, n > hoặc = 2 ta có
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
\(=\frac{3}{5}.\frac{5n-5}{45n+36}=\frac{n-1}{45n+36}\)
chứng minh rằng với mọi n thuộc N, n>= 2 thì
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}< \frac{1}{15}\)
Đặt A =\(\frac{3}{5}.\left(\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{\left(5n-1\right).\left(5n+4\right)}\right)\)
= \(\frac{3}{5}.\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)
= \(\frac{3}{5}.\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
= \(\frac{3}{5}.\frac{1}{9}-\frac{3}{5}.\frac{1}{5n+4}=\frac{1}{15}-\frac{3}{5.\left(5n+4\right)}< \frac{1}{15}\)( ĐPCM )
chứng minh rằng với mọi n thuộc N và n>=2 thì
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+...+\frac{3}{\left(5n+1\right)\left(5n+4\right)}< \frac{1}{15}\)
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+....+\frac{3}{\left(5n+1\right)\left(5n+4\right)}\)
\(=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+\frac{5}{19.24}+....+\frac{5}{\left(5n+1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+....+\frac{1}{5n+1}-\frac{1}{5n+4}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
\(=\frac{1}{15}-\frac{3}{5\left(5n+4\right)}< \frac{1}{15}\) (đpcm)
CMR : với mọi n thuộc N thì ta luôn có :
1/6+1/66+1/176+...+1/(5n+1)(5n+6)=n+1/5n+6