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F=14/8.24/18.36/30....10706/10700
7.2.8.3.9.4.10.5...106.101
= _________________________
8.1.9.2.10.3.11.4...107.100
(2.3.4.5...101).(7.8.9.10...106)
=______________________________
(8.9.10.11...107).(2.3.4.5...100)
101.7.
=_______=1
107
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=\(\frac{14\times24\times36\times..........\times10706}{8\times18\times32\times..........\times10700}\)
=\(\frac{2\times7\times3\times8\times.............\times101\times106}{1\times8\times2\times9\times...........\times100\times107}\)
=\(\frac{\left(2\times3\times...........\times101\right)\times\left(7\times8\times...........\times106\right)}{\left(1\times2\times........\times100\right)\times\left(8\times9\times............\times107\right)}\)
=707/107
\(E=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)
\(\Rightarrow E=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{15.16}\)
\(E=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(E=2.\frac{3}{16}=\frac{3}{8}\)
\(E=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(E=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)
\(E=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(E=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(E=2.\frac{3}{16}\)
\(E=\frac{3}{8}\)
2/
S = 2 + 22 + 23 +...+ 299
= (2+22+23) +...+ (297+298+299)
= 2(1+2+22)+...+297(1+2+22)
= 2.7 +...+ 297.7
= 7(2+...+297) chia hết cho 7
S = 2+22+23+...+299
= (2+22+23+24+25)+...+(295+296+297+298+299)
= 2(1+2+22+23+24)+...+295(1+2+22+23+24)
= 2.31+...+295.31
= 31(2+...+295) chia hết cho 31
3/
A = 1+5+52+....+5100 (1)
5A = 5+52+53+...+5101 (2)
Lấy (2) - (1) ta được
4A = 5101 - 1
A = \(\frac{5^{101}-1}{4}\)
4/
Đặt A là tên của biểu thức trên
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
........
\(\frac{1}{8^2}< \frac{1}{7.8}=\frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}< 1\)
Vậy...
5/
a, Gọi UCLN(n+1,2n+3) = d
Ta có : n+1 chia hết cho d => 2(n+1) chia hết cho d => 2n+2 chia hết cho d
2n+3 chia hết cho d
=> 2n+2 - (2n+3) chia hết cho d
=> -1 chia hết cho d => d = {-1;1}
Vậy...
b, Gọi UCLN(2n+3,4n+8) = d
Ta có: 2n+3 chia hết cho d => 2(2n+3) chia hết cho d => 4n+6 chia hết cho d
4n+8 chia hết cho d
=> 4n+6 - (4n+8) chia hết cho d
=> -2 chia hết cho d => d = {1;-1;2;-2}
Mà 2n+3 lẻ => d lẻ => d khác 2;-2 => d = {1;-1}
Vậy...
\(D=\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+\frac{1}{336}+\frac{1}{546}\)
\(D=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+\frac{1}{21.26}\)
\(D=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)
\(D=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{26}\right)\)
\(D=\frac{1}{5}\left(\frac{1}{1}-\frac{1}{26}\right)\)
\(D=\frac{1}{5}.\frac{25}{26}=\frac{5}{26}\)