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\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)
\(=\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{1.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{39}=\frac{5}{39}\)
2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)
=\(1-\frac{1}{15}=\frac{14}{15}\)
\(\Rightarrow S=\frac{7}{15}\)
a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15
A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)
A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)
A=2(1-1/15)
A=1/2.14/15
A=7/15
S=1/3 + 1/15 + 1/35 + 1/63 + 1/99
=>S=1/1*3+1/3*5+1/5*7+1/7*9+1/9*11
Nhân cả hai vế với 2 ta được:
=>2S=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11
=>2S=1-1/11
=>2S=11/11-1/11
=>2S=10/11
=>S=10/11 : 2
=>S=5/11
Vậy S= 5/11
\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(S=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(S=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(S=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(S=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
A=1/1*3+1/3*5+1/5*7+.....+1/99*101
A=1/3*(1-1/3+1/3-1/5+1/5-1/7+.......+1/99-1/101)
A=1/3*(1-1/101)
A=1/3*100/101
A=300/301
A = 1/5 + 1/13 + 1/14 + 1/15 + 1/60 + 1/61 + 1/62 + 1/63
Ta có : A = 1/5 + 1/13 + 1/14 + 1/15 + 1/60 + 1/61 + 1/62 + 1/63 < 1/5 + 1/12 + 1/12 + 1/12 + 1/60 + 1/60 + 1/60
= A < 1/5 + 1/4 + 1/20
= A < 1/2
Vậy A < 1/12
Ta có : \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)
Ta có:
\(S=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)
\(=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Bài toán phụ 1:
Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12x3=1/4 (1)
Bài toán phụ 2:
Ta có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60x3=1/20 (2)
Từ (1) và (2), ta có:
1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/5+1/4+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<4/20+5/20+1/20
1/5+1/13+1/14+1/15+1/61+1/62+1/63<9/20<1/2
=>1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
S = \(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{17x19}\)
2S = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\)\(\frac{1}{17}-\frac{1}{19}\)
2S = \(\frac{1}{3}-\frac{1}{19}\)
2S = \(\frac{16}{57}\)
S = \(\frac{16}{57}\times\frac{1}{2}\)
S = \(\frac{8}{57}\)
\(S=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}+\frac{1}{255}+\frac{1}{323}\)
\(S=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}\)
\(2S=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}\)
\(2S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\)
\(2S=\frac{1}{3}-\frac{1}{19}\)
\(2S=\frac{19}{57}-\frac{3}{57}\)
\(2S=\frac{16}{57}\)
\(S=\frac{16}{57}:2\)
\(S=\frac{16}{57}\cdot\frac{1}{2}\)
\(S=\frac{8}{57}\)
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}\times\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\times\frac{12}{13}\)
\(=\frac{6}{13}\)
1/3 + 1/3×5 + 1/5×7 + 1/7×9 + 1/9×11 + 1/11×13
=> 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 +1/9 -1/11 + 1/11 + -1/13
=> 2/3 - 1/13 = 23/39
Đúng 100% nha. k mk nha
\(S=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{37.39}\)
=> \(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{37.39}\right)\)
=> \(S=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{37}-\frac{1}{39}\right)\)
=> \(S=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{39}\right)=\frac{2}{13}\).