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Ta có :
\(A=\frac{101}{1}+\frac{100}{2}+\frac{99}{3}+...+\frac{1}{101}\)
\(A=\left(101-1-...-1\right)+\left(\frac{100}{2}+1\right)+\left(\frac{99}{3}+1\right)+...+\left(\frac{1}{101}+1\right)\)
\(A=\frac{102}{102}+\frac{102}{2}+\frac{102}{3}+...+\frac{102}{101}\)
\(A=102\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}+\frac{1}{102}\right)\)
\(\Rightarrow\)\(\frac{A}{B}=\frac{102\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{102}}=\frac{102}{1}=102\)
Vậy \(\frac{A}{B}=102\)
Chúc bạn học tốt ~
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 )
Câu 1:
Đặt: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+....+\frac{1}{100^2}\)
\(=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+....+\frac{1}{100.100}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow A< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy:.............
Câu 2:
\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{98}+1\right)\left(\frac{1}{99}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{98}+\frac{98}{98}\right)\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{99}{98}.\frac{100}{99}\)
\(=\frac{3.4.5....99.100}{2.3.4...98.99}\)
\(=\frac{100}{2}=50\)
Mk làm mẫu câu a nha
a, Có :
2A = 1+1/2+1/2^2+.....+1/2^98
A = 2A - A = (1+1/2+1/2^2+.....+1/2^98)-(1/2+1/2^2+......+1/2^99) = 1 - 1/2^99
Tk mk nha
nhân A với 2 rồi lấy 2A-A