\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\right)\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< 2\)
\(\Rightarrow A< \frac{1}{2^2}.2=\frac{1}{2}\) (đpcm)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{2^2-1}+\frac{1}{4^2-1}+\frac{1}{6^2-1}+...+\frac{1}{100^2-1}\)
\(A< \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(2A< \frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A< \frac{1}{1}-\frac{1}{101}< 1\Rightarrow2A< 1\Rightarrow A< \frac{1}{2}\left(đpcm\right)\)
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(2A=\frac{2}{2^2}+\frac{2}{4^2}+\frac{2}{6^2}+...+\frac{2}{100^2}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
gọi B là tên biểu thức vế phải
lại có :
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
\(B=1-\frac{1}{2^{50}}< 1\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \frac{1}{2}\)
