chứng minh rằng với mọi số nguyên n , n \(\ge\) 2, ta có:\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+.......+\frac{3}{\left(5n-1\right)\left(5n+4\right)}< \frac{1}{15}\)
Chứng minh rằng với mọi n thuộc N; n nhỏ hơn hoặc bằng 2 ta có:
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+....+\frac{3}{\left(5n-1\right).\left(5n+4\right)}<\frac{1}{15}\)
Đặt \(A=\frac{3}{9.14}+\frac{3}{14.19}+.......+\frac{3}{\left(5n-1\right)\left(5n+4\right)}\)
\(5A=\frac{15}{9.14}+\frac{15}{14.19}+.....+\frac{15}{\left(5n-1\right)\left(5n+4\right)}\)
\(5A=3.\left(\frac{5}{9.14}+\frac{5}{14.19}+......+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(5A=3.\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)
\(5A=3.\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
\(5A=\frac{1}{3}-\frac{1}{5n+4}\)
=> \(5A<\frac{1}{3}\)
=> \(A<\frac{1}{3}:5\)
hay \(A<\frac{1}{15}\) \(\left(đpcm\right)\)
Nhớ nhé bạn
Chứng minh rằng với mọi n thuộc N ; n nhỏ hơn hoặc bằng 2 ta có:
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+......+\frac{3}{\left(5n-1\right).\left(5n+4\right)}<\frac{1}{15}\)
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)
chứng minh rằng : với n thuộc N ; n>1
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+.....+\frac{3}{\left(5n-1\right)\left(5n+4\right)}<\frac{1}{15}\)
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 Với n \(\in\) N; n \(\ge\) Z, ta có:
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+...+\frac{3}{\left(5n-1\right)\left(5n-4\right)}
\(\frac{3}{9}\)- \(\frac{3}{14}\)+ \(\frac{3}{14}-\frac{3}{19}+\frac{3}{19}-\frac{3}{24}+...+\frac{3}{5n-1}-\frac{3}{5n-4}=\frac{3}{9}-\frac{3}{5n-4}=\frac{3\left(5n-4\right)}{9\left(5n-4\right)}-\frac{27}{9\left(5n-4\right)}=\frac{15n-12-27}{45n-36}=\frac{15n-39}{45n-36}\)
\(\frac{15n-39}{45n-36};\frac{1}{5}\)
so sanh
\(\frac{\left(15n-39\right)5}{\left(45n-36\right)5}=\frac{75n-195}{225n-180}\)
\(\frac{1}{5}=\frac{45n-36}{5\left(45n-36\right)}=\frac{45n-36}{225n-180}\)
vì 75n-195 < 45n-36 suy ra dãy số trên bé hơn 1/5
Chứng minh với mọi \(n\in N;n\ge2\)ta có :
\(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n-1\right)\left(5n+4\right)}<\frac{1}{15}\)
\(CMR:\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+5+...+59}< \frac{2}{3}\)
\(\frac{3}{9.14}+\frac{3}{14.19}+\frac{3}{19.24}+...+\frac{3}{\left(5n-1\right).\left(5n+4\right)}< \frac{1}{15}\)