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\(A=3+3^2+...+3^{50}\)
\(\Rightarrow3A=3^2+3^3+...+3^{50}+3^{51}\)
\(\Rightarrow3A-A=3^{51}-3\)
\(\Rightarrow2A=3^{51}-3\)
\(\Rightarrow A=\frac{3^{51}-3}{2}\)
\(B=2-2^2+2^3-2^4+...+2^{2019}-2^{2020}\)
\(2B=2^2-2^3+2^4-2^5+...+2^{2020}-2^{2021}\)
\(B+2B=2-2^{2021}\)
\(3B=2-2^{2021}\)
\(B=\frac{2-2^{2021}}{3}\)
\(C=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2008.2009}\)
\(C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(C=1-\frac{1}{2009}\)
\(C=\frac{2008}{2009}\)
\(D=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(D=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(D=\frac{1}{2}\left(1-\frac{1}{11}\right)\)
\(D=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)
1. a) 2B = 1 + 1/2 + 1/22+...+1/298
B - B = (1+1/2+...+1/298) - (1/2+....+1/299)
B = 1 - 299 => B < 1
b) Làm tương tự như câu a, ra là (1 - 1/399) : 2 = 1/2 - 1/2.399(C bé hơh 1/2)
1. a). Theo đầu bài ta có:
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\Leftrightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Leftrightarrow B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{2^{99}}< 1\)( đpcm )
\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
Bài 1:
C = 1/101 + 1/102 + 1/103 + ... + 1/200
Có:
C < 1/101 + 1/101 + 1/101 + ... + 1/101
C < 100 . 1/101
C < 100/101
Mà 100/101 < 1
=> C < 1 (1)
Có:
C > 1/200 + 1/200 + 1/200 + ... + 1/200
C > 100 . 1/200
C > 1/2 (2)
Từ (1) và (2)
=> 1/2<C<1
Ủng hộ nha mk làm tiếp
\(\frac{-9}{4}\).\(19\frac{2}{5}\)+\(\left(\frac{-3}{2}\right)^2\).\(\left(-14\frac{3}{5}\right)\)-\(\left(\frac{99}{100}\right)^0\)
=\(\frac{-9}{4}\).\(\frac{97}{5}\)+\(\frac{9}{4}\).\(\frac{-73}{5}\)-1
=\(\frac{-9}{4}\).\(\frac{97}{5}\)+\(\frac{-9}{4}\).\(\frac{73}{5}\)-1
=\(\frac{-9}{4}\).(\(\frac{97}{5}\)+\(\frac{73}{5}\))
=\(\frac{-9}{4}\).34
=\(\frac{-153}{2}\)
Học tốt
Đặt A=\(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\)
3A=\(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^{98}}\)
2A = 3A - A = \(\frac{1}{3}-\frac{1}{3^{98}}\)<\(\frac{1}{2}\)
=> A = \(\frac{\frac{1}{3}-\frac{1}{3^{98}}}{2}<\frac{1}{2}\)(đpcm)
Đặt B=2100-299+298-297+.....+22-2
2A=2101-2100+299-298+...+23-22
3A=2A+A=2101-2
=> A=\(\frac{2^{101}-2}{3}\)
pk ko ban