\(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+....+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
Những câu hỏi liên quan
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)
Chứng minh rằng với mọi n \(\in\) N ta luôn có:
\(\dfrac{1}{1.6}\)+\(\dfrac{1}{6.11}\)+\(\dfrac{1}{11.16}\)+...+\(\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}\)= \(\dfrac{n+1}{5n+6}\)
\(\dfrac{1}{1\cdot6}+\dfrac{1}{6\cdot11}+\dfrac{1}{11\cdot16}+...+\dfrac{1}{\left(5n+1\right)\left(5n+6\right)}=\dfrac{n+1}{5n+6}\)
\(VT=\dfrac{1}{5}\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{\left(5n+1\right)\left(5n+6\right)}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\left(1-\dfrac{1}{5n+6}\right)\)\(=\dfrac{1}{5}\cdot\left(\dfrac{5n+6}{5n+6}-\dfrac{1}{5n+6}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{5\left(n+1\right)}{5n+6}=\dfrac{n+1}{5n+6}=VP\)
Đúng 0
Bình luận (4)
Ta có: \(\dfrac{1}{1.6}+\dfrac{1}{6.11}+\dfrac{1}{11.16}+...+\dfrac{1}{\left(5n+1\right).\left(5n+6\right)}\)
=\(\dfrac{1}{5}.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{5n+1}-\dfrac{1}{5n+6}\right)\)
=\(\dfrac{1}{5}.\left(1-\dfrac{1}{5n+6}\right)\)
= \(\dfrac{1}{5}.\left(\dfrac{5n+6}{5n+6}-\dfrac{1}{5n+6}\right)\)
=\(\dfrac{1}{5}.\dfrac{5n+5}{5n+6}\)
=\(\dfrac{1}{5}.\dfrac{5.\left(n+1\right)}{5n+6}\)
=\(\dfrac{n+1}{5n+6}\left(ĐPCM\right)\)
Đúng 0
Bình luận (0)
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!
Đúng 0
Bình luận (0)
Chứng minh:
c, \(\dfrac{3}{9.14}+\dfrac{3}{14.19}+\dfrac{3}{19.24}+....+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}< \dfrac{1}{15}\)
Đặt :
\(A=\dfrac{3}{9.14}+\dfrac{3}{14.19}+........+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{5}{9.14}+\dfrac{5}{14.19}+........+\dfrac{5}{\left(5n-1\right)\left(5n+4\right)}\)
\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...........+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\)
\(\Leftrightarrow\dfrac{5}{3}A=\dfrac{1}{9}-\dfrac{1}{5n+4}\)
\(\Leftrightarrow A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right):\dfrac{5}{3}\)
\(\Leftrightarrow A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right).\dfrac{3}{5}\)
\(\Leftrightarrow A=\dfrac{1}{9}.\dfrac{3}{5}-\dfrac{1}{5n+4}.\dfrac{3}{5}\)
\(\Leftrightarrow A=\dfrac{1}{15}-\dfrac{1}{5n+4}.\dfrac{3}{5}< \dfrac{1}{15}\)
\(\Leftrightarrow A< \dfrac{1}{15}\left(đpcm\right)\)
Đúng 0
Bình luận (0)
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)
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
Chứng minh rằng với mọi \(n\in N\); \(n\ge2\) ta có :
\(\dfrac{3}{9.14}+\dfrac{3}{14.19}+....................+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}< \dfrac{1}{15}\)
Đặt :
\(A=\dfrac{3}{9.14}+\dfrac{3}{14.19}+......................+\dfrac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(A.\dfrac{5}{3}=\dfrac{5}{9.14}+\dfrac{5}{14.19}+..................+\dfrac{5}{\left(5n-1\right)\left(5n+1\right)}\)
\(A.\dfrac{5}{3}=\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+..................+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\)
\(A.\dfrac{5}{3}=\dfrac{1}{9}-\dfrac{1}{5n+4}\)
\(A=\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right):\dfrac{3}{5}\)
\(A=\left(\dfrac{1}{9}-\dfrac{1}{5n+\text{4}}\right).\dfrac{3}{5}\)
\(A=\dfrac{1}{9}.\dfrac{3}{5}-\dfrac{1}{5n+4}.\dfrac{3}{5}\)
\(A=\dfrac{1}{15}-\dfrac{1}{5.\left(5n+4\right)}\)
\(\Rightarrow A< \dfrac{1}{15}\)
\(\Rightarrowđpcm\)
Chúc bn học tốt!!!!!!!!!!
Đúng 0
Bình luận (0)
Tính :6/ lim\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
7/ lim \(\dfrac{\sqrt{n^3-2n+5}}{3+5n}\)
10/ lim\(\dfrac{1+3+5+...+\left(2n+1\right)}{3n^3+4}\)