Question

Cho C = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}.\).Chứng minh rằng C<2

Answer 1

TA CÓ Vế trái <\(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)

\(=2-\frac{1}{50}< 2\)

do đó VT <2(dpcm)

Answer 2

1+12=13

Answer 3

\(\frac{1}{1^2}=1\)

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

  \(.\)         \(.\)

  \(.\)         \(.\)

  \(.\)         \(.\)

\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)

Cộng vế với vế ta có \(:\)

\(C< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...........+\frac{1}{49\cdot50}\)

\(C< 1+\frac{49}{50}< \frac{50+49}{50}=\frac{99}{50}< \frac{100}{50}=2\)

Vậy \(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}< 2\)

Answer 4

\(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.......+\frac{1}{49\cdot50}\)

\(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{50^2}< 1+1-\frac{1}{50}\)

\(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}< 2-\frac{1}{50}< 2\)

\(\)Vậy \(C< 2\)( điều cần chứng minh )