Question

1.RÚT GỌN: B= 3¹⁰⁰-3⁹⁹+3⁹⁸-3⁹⁷+...+3²-3+1

2.RÚT GỌN: C= 1/3 + 1/3²  + 1/3³ +...+1/3⁹⁹ . CMR C < 1/2

(Nhờ các bn giúp mk nhé sắp phải nộp rùi)

Answer 1

\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)

\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)

\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)

\(2C=1-\frac{1}{3^{99}}< 1\)

\(\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)

Answer 2

1.

B = 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1

3B = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3

3B + B = ( 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3 ) + ( 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1 )

4B = 3¹⁰¹ + 1

B = \(\frac{3^{101}+1}{4}\)

Answer 3

1.   B = 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ - 3⁹⁷ +... + 3² - 3 + 1

     3B = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹- 3⁹⁸ +... + 3³ - 3² + 3

3B + B = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸ + ... + 3³ - 3² + 3 + 3¹⁰⁰ - 3⁹⁹ + 3⁹⁸ + ... + 3² - 3 + 1

      4B = 3¹⁰¹ + 1

         B = \(\frac{3^{101}+1}{4}\)

2. Ta có:

C = 1/3 + 1/3² + 1/3³ + ... + 1/3⁹⁹

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

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

\(\frac{2C}{3}=\frac{1}{3}-\frac{1}{3^{100}}\)

\(2C=1-\frac{1}{3^{99}}\)

  \(C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)

Vậy : C < \(\frac{1}{2}\)( đpcm ) .