Question

Bai 1; a) Cho \(B\) = \(\frac{1}{2}\)+   (\(\frac{1}{2}\))²  +     (\(\frac{1}{2}\))³  +   (\(\frac{1}{2}\))⁴  + ... +  (\(\frac{1}{2}\))⁹⁸  +   (\(\frac{1}{2}\))⁹⁹ . Chứng minh rằng  \(B\) \(< 1\)

          b) Cho \(C\) = \(\frac{1}{3}\)+    \(\frac{1}{3^2}\)+   \(\frac{1}{3^3}\)+...+   \(\frac{1}{3^{98}}\)+   \(\frac{1}{3^{99}}\).  Chứng minh rằng \(C\) \(< \)\(\frac{1}{2}\)

Bai 2;Chứng minh rằng;

           a) \(\frac{3}{1^2.2^2}\)+   \(\frac{5}{2^2.3^2}\)+   \(\frac{7}{3^2.4^2}\)+ ... +   \(\frac{19}{9^2.10^2}\)\(< \)\(1\)

           b) \(\frac{1}{3}\)+  \(\frac{2}{3^2}\)+\(\frac{3}{3^3}\)+ ... +  \(\frac{99}{3^{99}}\)+   \(\frac{100}{3^{100}}\)\(< \frac{3}{4}\)

Answer 1

1. a) 2B = 1 + 1/2 + 1/2²+...+1/2⁹⁸

B - B = (1+1/2+...+1/2⁹⁸) - (1/2+....+1/2⁹⁹)

B = 1 - 2⁹⁹ => B < 1

b) Làm tương tự như câu a, ra là (1 - 1/3⁹⁹) : 2 = 1/2 - 1/2.3⁹⁹(C bé hơh 1/2)

Answer 2

1. a). Theo đầu bài ta có:
 \(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\Leftrightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Leftrightarrow B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{2^{99}}< 1\)( đpcm )

Answer 3

1. b). Theo đầu bài ta có:
\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(\Leftrightarrow C=\frac{\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)}{2}\)
\(\Leftrightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)( đpcm )

Answer 4

Bài 2b. Theo đầu bài ta có:
\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\)
\(=\frac{\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{99}{3^{98}}+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}\right)}{2}\)
\(=\frac{\left(1-\frac{100}{3^{100}}\right)+\left(\frac{2}{3}-\frac{1}{3}\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+...+\left(\frac{99}{3^{98}}-\frac{98}{3^{98}}\right)+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)}{2}\)
\(=\frac{\left(1-\frac{100}{3^{100}}\right)+\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)}{2}< \frac{1+\frac{1}{2}}{2}=\frac{3}{2}:2=\frac{3}{4}\)
\(\Rightarrow\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{99}{3^{99}}+\frac{100}{3^{100}}< \frac{3}{4}\) ( đpcm )

Answer 5

Mọi người làm 1a,1b,2b rồi để mình giúp bạn bài 2a thôi nhé:

Cái này có quy luật cả rồi

3/1^2.2^2  +  5/2^2.3^2  +  .... +  19/9^2.10^2

=3/1.4  +  5/4.9  +  7/9.16  +  .... +  19/81.100

=1/1 - 1/4  +  1/4  -  1/9  +  1.9  +  1/16  + .... + 1/81  -  1/100

=1/1-1/100 <1 (đpcm)

Answer 6

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

               \(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)

              \(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

          Đặt \(B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

                \(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

               \(\Rightarrow2B=3-\frac{1}{3^{99}}\)

              \(\Rightarrow2B< 3\Rightarrow B< \frac{3}{2}\)

        Vì \(2A=B-\frac{100}{3^{100}}\Rightarrow2A< B\)

           \(\Rightarrow2A< \frac{3}{2}\Rightarrow A< \frac{3}{4}\)(đpcm)