Question

Tìm x biết

a 1/1.2 + 1/2.3 +...+ 1/x.(x+1) = 44/45

b Chứng tỏ rằng: 1/2^2 + 1/3^2 + ... + 1/45^2 < 1

Answer 1

a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)

=> \(1-\frac{1}{x+1}=\frac{44}{45}\)

=> \(\frac{x}{x+1}=\frac{44}{45}\)

=> x = 44

b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

.................

\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)

Answer 2

a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45

=> x=44

b/ 1/2² < 1/1.2; 1/3² < 1/2.3; ....; 1/45² < 1/44.45

=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1

=> A < 1