Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{2}M=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(M-\frac{1}{2}M=\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^3}+\frac{1}{2^3}\right)+...+\frac{1}{2}-\frac{1}{2^{100}}\)
\(M=\left(\frac{1}{2}-\frac{1}{2^{100}}\right).2=1-\frac{1}{2^{10000}}\)
Vậy M < 1
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
Mà
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{3-2}{2.3}+\frac{4-3}{3.4}\frac{5-4}{4.5}+...+\frac{12-11}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)
1/2^2+1/3^2+1/4^2+....+1/11^2<1/(2.3)+1/(3.4)+1/(4.5)+.....+1/(11.12)
=1/2-13+1/3-1/4+1/5+.....+1/11-1/12
=1/2-1/12=5/12
VẬY A<5/12
ks cho mình nhé
\(M=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+....+\frac{1}{2^{43}}-\frac{1}{2^{46}}+\frac{1}{2^{49}}-\frac{1}{2^{52}}\)
Nên \(2^3.M=4-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+.....+\frac{1}{2^{46}}-\frac{1}{2^{52}}\)
Suy ra \(2^3.M-M=4-\frac{1}{2^{52}}\)hay\(7.M=4-\frac{1}{2^{52}}\).
Khi đó \(M=\frac{4}{7}-\frac{1}{2^{52}.7}< 1\)
Vì \(\frac{9}{4}>1;M< 1\)nên \(\frac{9}{4}>M\)
Vậy \(\frac{9}{4}>M\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+...+\frac{1}{3^{2013}}\)
\(\Rightarrow3A-A\)= \(\left(3+1+...+\frac{1}{3^{2013}}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^{2014}}\right)\)
\(\Rightarrow2A=3-\frac{1}{3^{2014}}\)
\(\Rightarrow A=\frac{3-\frac{1}{3^{2014}}}{2}\)
\(\Rightarrow A=\frac{3}{2}-\frac{\frac{1}{3^{2014}}}{2}< \frac{3}{2}\)
Vậy \(A< \frac{3}{2}\)
Chúc bạn học tốt !!!
Mình làm bài 2 nhé:
Ta có: \(\frac{1}{2^2}<\frac{1}{2\times3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3^2}<\frac{1}{3\times4}=\frac{1}{3}-\frac{1}{4}\)
....
\(\frac{1}{50^2}<\frac{1}{50\times51}=\frac{1}{50}-\frac{1}{51}\)
Tổng các vế ta sẽ có \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}<\frac{1}{2}-\frac{1}{51}=\frac{49}{102}<1\)