Question

chứng minh rằng :

a) 1+3+3^2+3^3+...+3^11 chia hết cho 40

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

Answer 1

a) 1 + 3 + 3² + 3³ + ... + 3¹¹

= (1 + 3 + 3² + 3³) + ... + (3⁸ + 3⁹ + 3¹⁰ + 3¹¹)

= 40 + ... + 3⁸.(1 + 3 + 3² + 3³)

= 40 + ... + 3⁸. 40

= (1 + ... + 3⁸) . 40 \(⋮\)40

Answer 2

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

        =>    B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

       => B < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

      => B <\(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)

     => B < \(1-\frac{1}{100}\)

    => B < 1

Answer 3

a) \(1+3+3^2+3^3+...+3^{11}\)

\(=\left(1+3+3^2+3^3\right)+...+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(=40.1+40.3^4+40.3^8\)

\(=40.\left(1+3^4+3^8\right)\)

=> BT chia hết cho 40

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

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1-\frac{1}{100}< 2=1\) (ĐPCM)