Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
gọi biểu thức cần CM là B
\(3B=1+\frac23+\frac{3}{2^2}+\frac{4}{3^3}+\cdots+\frac{100}{3^{99}}\)
=> \(3B-B=1+\left(\frac23-\frac13\right)+\left(\frac{3}{3^2}-\frac{2}{3^2}\right)+\cdots+\left(\frac{100}{3^{99}}-\frac{99}{3^{99}}\right)-\frac{100}{3^{100}}\)
\(2B=1+\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
đặt C= \(\frac13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^{99}}\)
=> \(3C=1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\)
=> \(3C-C=\left(1+\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{98}}\right)-\left(\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}\)
=> \(C=\frac12-\frac{1}{2\cdot3^{99}}\)
\(2B=1+\frac12-\left(\frac{1}{2\cdot3^{99}}+\frac{100}{3^{100}}\right)\)
vì trong ngoặc lớn hơn 0
=> \(2B<\frac32\)
\(B<\frac34\left(đpcm\right)\)
\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\left(\frac{1}{4\cdot5}+.\ldots+\frac{1}{99\cdot100}\right)\)
\(A=\frac{7}{12}+\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)\)
Mà \(\left(\frac{1}{4\cdot5}+\cdots+\frac{1}{99\cdot100}\right)>0\)
=> A>\(\frac{7}{12}\)
mặt khác ta có: \(A=1-\frac12+\frac13-\frac14+\frac14-\frac15+\cdots+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\left(\frac12-\frac13\right)-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)
\(A=\frac56-\left(\frac14-\frac15\right)-.\ldots-\left(\frac{1}{98}-\frac{1}{98}\right)-\frac{1}{100}\)
=> \(A<\frac56\)
Vậy \(\frac{7}{12}<A<\frac56\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{99}{100}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
a,1/102+1/112+1/122+...+1/1002<1/9.10+1/10.11+1/11.12+...+1/99.100=1/9-1/10+1/10-1/11+...+1/99-1/100
=1/9-1/100=91/900<3/4
Vậy 1/102+1/112+1/122+...+1/1002<3/4
b,1/22+1/32+1/42+...+1/1002<1/1.2+1/2.3+1/3.4+...+1/99.100=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
=1-1/100=99/100
Vậy 1/22+1/32+1/42+...+1/1002<99/100
c,1/22+1/32+1/42+...+1/1002<1/22+(1/2.3+1/3.3+...+1/99.100)=1/4+(1/2-1/3+1/3-1/4+...+1/99-1/100)
=1/4+(1/2-1/100)=1/4+49/100=74/100<3/4=75/100
Vậy 1/22+1/32+1/42+...+1/1002<3/4
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~