Question

[Bùi Bảo Ngọc]

Chứng minh rằng

a,1/2²+1/4²+1/6²+...+1/100²<1/2

b,1/3+1/31+1/37+1/47+1/53+1/61<1/2

3/4+8/9+15/16+..+2499/2500>48

Answer 1

[Thanh Tùng DZ]

a) gọi \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)

gọi \(B=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)

\(B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1+1-\frac{1}{50}\)

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

\(\Rightarrow A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)

b) Ta thấy \(\frac{1}{37}< \frac{1}{35}< \frac{1}{31}< \frac{1}{30}\), \(\frac{1}{61}< \frac{1}{53}< \frac{1}{47}< \frac{1}{45}\)

Do đó : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}.3+\frac{1}{45}.3=\frac{1}{2}\)

c) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)

\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)

\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)

\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)

Ta thấy vế trong ngoặc nhỏ hơn 1

\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>48\)