Question

[Luger Girl]

a} Tinh M : N biet M = 1/11.13 + 1/13.15 + 1/15.17 +........+1/33.35 va M = 12/11.13.15 + 12/13.15.17 + 12/15.17.19 +.....+12/31.33.35

b} Cho S = 1 - 1/2 + 1/3 -1/4 +.......+ 1/2011 - 1/2012 + 1/2013 va P = 1/1007 + 1/1008 +.......+ 1/2012 + 1/2013 . Tinh [S - P - 1]

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Answer 1

[ST]

a, Ta có: 

\(M=\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{33.35}\)

\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{33}-\frac{1}{35}\right)\)

\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{35}\right)=\frac{1}{2}\cdot\frac{24}{385}=\frac{12}{385}\)

\(N=\frac{12}{11.13.15}+\frac{12}{13.15.17}+...+\frac{12}{31.33.35}\)

\(=3\left(\frac{1}{11.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.17}+...+\frac{1}{31.33}-\frac{1}{33.35}\right)\)

\(=3\left(\frac{1}{11.13}-\frac{1}{33.35}\right)=3\cdot\frac{92}{15015}=\frac{92}{5005}\)

\(\Rightarrow M:N=\frac{12}{385}:\frac{92}{5005}=\frac{39}{23}\)

b, \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)

\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)

\(\Rightarrow S-P-1=\left(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\right)-\left(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\right)-1=0-1=-1\)

Answer 2

[Luger Girl]

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