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\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};......;\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-1\right)n}\)
Ta lại có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+.....+\frac{1}{n\left(n-1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{n^2}< 1\) (đpcm)
Cho \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
so sánh B với \(\frac{3}{4}\)
Ta có:\(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\)
B < \(\frac{1}{4}\) < \(\frac{3}{4}\)
\(\Leftrightarrow B< \frac{3}{4}\)
Ta có:
a=\(\left(\frac{-1}{2^2}\right).\left(\frac{-1}{3^2}\right)......\left(\frac{-1}{100^2}\right)=\left(\frac{\left(-1\right).\left(-1\right).......\left(-1\right)}{2^2.3^2........100^2}\right)\)
Vì có 98 phân số
=> có 98 số -1 nhân với nhau
=> tích của 98 số -1 =1 vì số số hạng của nó là số chẵn
=>\(\left(\frac{-1}{2^2}\right).\left(\frac{-1}{3^2}\right)......\left(\frac{-1}{100^2}\right)=\left(\frac{\left(-1\right).\left(-1\right).......\left(-1\right)}{2^2.3^2........100^2}\right)\)
=\(\frac{1}{2^2.3^2.......100^2}>0\)
mà \(\frac{-1}{2}< 0\)
=>\(\frac{-1}{2}< \frac{1}{2^2.3^2.............100^2}\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)(99 số hạng)
\(\Rightarrow A=\left(\frac{-3}{4}\right)\left(\frac{-8}{9}\right)...\left(\frac{-9999}{10000}\right)\)
\(\Rightarrow-A=\frac{3}{4}.\frac{8}{9}...\frac{9999}{10000}\)
\(\Rightarrow-A=\frac{1.3.2.4....99.101}{2.2.3.3.4.4...100.100}\)
\(\Rightarrow-A=\frac{1.2.3...99}{2.3...100}.\frac{2.3.4...101}{2.3.4...100}\)
\(\Rightarrow-A=\frac{1}{100}.101=\frac{101}{100}\)
\(\Rightarrow A=-\frac{101}{100}< -\frac{50}{100}=-\frac{1}{2}\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{56}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{55}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{55}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{56}}\right)\)
\(A=1-\frac{1}{2^{56}}< 1\)