Question

[ミ★ηɠυүễη ʋăη ɦưηɠ ²ƙ⁸★彡⁀ᶦᵈᵒᶫ]

A=\(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)                 CMR:A<\(\frac{1}{16}\)

Answer 1

[Xyz OLM]

Ta có : A = \(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)

=> 5A = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)

=> 5A - A =  \(\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\right)\)

=> 4A \(=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=> 20A = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)

Lấy 20A trừ A ta có : 

20A - A = \(\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)

16A = \(1-\frac{99}{5^{99}}+\frac{99}{5^{100}}=1+99\left(\frac{1}{5^{100}}-\frac{1}{5^{99}}\right)=1-\frac{99.4}{5^{100}}\)

=> A = \(\frac{1}{16}-\frac{99}{4.5^{100}}< \frac{1}{16}\left(\text{ĐPCM}\right)\)

Answer 2

[I - Vy Nguyễn]

Ta có :A=\(\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)

          5A=\(\frac{1}{5}+\frac{2}{5^2}+.....+\frac{99}{5^{99}}\)

      5A -A=\(\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{99}{5^{99}}\right)\)-\(\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)

         4A  =\(\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Đặt B=\(\frac{1}{5}+\frac{1}{5^2}+.....+\frac{1}{5^{99}}\)

         5B=\(1+\frac{1}{5}+...+\frac{1}{5^{98}}\)

  5B - B =\(\left(1+\frac{1}{5}+...+\frac{1}{5^{98}}\right)\)- \(\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)

      4B  =\(1-\frac{1}{5^{99}}\)

 Ta có :4A = B -\(\frac{99}{5^{100}}\)

          16A = 4B -\(\frac{4.99}{5^{100}}\)=\(1-\frac{1}{5^{99}}-\frac{4.99}{5^{100}}\)

              A = \(\frac{1}{16}-\frac{1}{5^{99}.16}-\frac{99}{5^{100}.4}\)< \(\frac{1}{16}\)  

              Suy ra: A <\(\frac{1}{16}\)