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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; A = 1 + 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 2
1 = 1 = 1
1/2^2 < 1/1.2 = 1/1 - 1/2
1/3^2 < 1/2.3 = 1/2 - 1/3
.......................
1/100^2 < 1/99.100 = 1/99 - 1/100
Cộng vế với vế ta có:
A = 1 + 1 - 1/100
A = 2 - 1/100 < 2 (đpcm)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}< 2\left(đpcm\right)\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) 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 ~
Ta có\(\frac{-2}{3}\)+\(\frac{1}{4}\)= \(\frac{-8}{12}\)+\(\frac{3}{12}\)= \(\frac{-5}{12}\)
\(\frac{3}{4}\)-\(\frac{1}{3}\)=\(\frac{9}{12}\)-\(\frac{4}{12}\)=\(\frac{5}{12}\)
=> \(\frac{-5}{12}\)<\(\frac{a}{6}\)<\(\frac{5}{12}\)
=> \(\frac{-5}{12}\)<\(\frac{2a}{12}\)<\(\frac{5}{12}\)
Mà a là số nguyên,2a là số chẵn
=>2a
{-4,-2,0,2,4}
=>a
{-2,-1,0,1,2}