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2 vế bằng nhau
100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
100- 1-1/2-1/3-...-1/100 = 1/2+2/3+3/4+...+99/100
100 = 1 + 1/2 + 1/2 + 1/3 + 2/3 + ... + 1/100 + 99/100 (cùng cộng 2 vế với (- 1-1/2-1/3-...-1/100)
100 = 1 + 1 + 1 + ... + 1 (100 số hạng)
100 = 100
Vậy 100-(1+1/2+1/3+...+1/100) = 1/2+2/3+3/4+...+99/100
\(P=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3P=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3P-P=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)\(2P=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6P=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6P-2P=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4P=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4P=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4P=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4P=3-\frac{203}{3^{100}}< 3\)
\(P< \frac{3}{4}\)
\(P< Q\)
#)Giải :
\(A=\frac{1}{3^1}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{1}{3^{100}}\)
\(A=\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\)
\(\Rightarrow2A=1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\)
\(\Rightarrow2A-A=A=\left(1+\frac{2}{9}+\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{49}}\right)-\left(\frac{2}{9^1}+\frac{2}{9^2}+\frac{2}{9^3}+...+\frac{2}{9^{50}}\right)\)
\(\Rightarrow A=1+\frac{2}{9}-\frac{2}{9^{50}}=\frac{11}{9}-\frac{2}{9^{50}}\)
Có lẽ đúng .........................
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)