1.So sánh: A=\(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\) và 1:
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1.So sánh: A=\(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\) và 1:
ta có 1/2^2<1/2
1/2^3<1/2
.............
1/2^50<1/2
\(\Rightarrow\)1/2*50>1/2^1+1/2^2+1/2^3+...........+1/2^50
\(\Rightarrow\)
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1.So sánh: A=\(\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\) và 1:
\(A=\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}\)
\(\Rightarrow A< 1\)
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So sánh :
\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)và 1
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}< 1\)
\(\Rightarrow A< 1\)
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\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\) và\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\) so sánh A và B
ChoA=1/26+1/27+1/28+.. +1/49, B=1-1/2+1/3-1/4+... +1/49-1/50
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B = 1 + 1/2 + 1/3 + ... + 1/50
b = (1 + 1/3 + 1/5 + ... + 1/49) + (1/2 + 1/4 + 1/6 + ... + 1/50)
b = (1 + 1/2 + 1/3 + 1/4 + ... + 1/50) - 2(1/2 + 1/4 + 1/6 + ... + 1/50)
b = 1 + 1/2 + ... + 1/50 - 1 - 1/2 - 1/3 - ... - 1/25
b = 1/26 + 1/27 + 1/28 + ... + 1/50
vậy a = b
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so sánh \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{2}{2^{49}}+\frac{2}{2^{50}}\)với 1
2A=1+1/2+................+1/2^49+1/2^50
A=1+1/2^50=> A>1
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Giúp mình với:
So sánh A và B biết:
\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
Cảm ơn các bạn :)
ta có : \(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{50}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)\)
\(B=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
\(\Rightarrow\)\(B=A\)
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Tính và so sánh: \(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}...+\frac{99}{49^2.50^2}\)\(T=\frac{1}{2^2-1^2}+\frac{1}{3^2-1^2}+\frac{1}{4^2-1^2}+...+\frac{1}{50^2-1^2}\)
\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{49^2}-\frac{1}{50^2}\)
\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)
\(T=\frac{1}{\left(2-1\right)\left(2+1\right)}+\frac{1}{\left(3-1\right)\left(3+1\right)}+...+\frac{1}{\left(50-1\right)\left(50+1\right)}\)
\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)
Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T
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JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)
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Bài 1 :Chứng tỏ rằng :
\(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}\)\(-\frac{5}{3}+\frac{3}{2}-1\)
Bài 2 : Cho
\(A=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{4998}{4999}\)
Hãy so sánh A và 0,02
Câu hỏi của Lê Thị Minh Trang - Toán lớp 6 - Học toán với OnlineMath
Xem bài 1 nhé !
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Bài 1:
Xét vế phải :
\(P=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}\)\(-1=2\)\(\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left(\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}\)
Đẳng thức được chứng tỏ là đúng
Bài 2 :
Đặt \(A'=\frac{3}{4}.\frac{4}{5}.\frac{7}{8}...\frac{4999}{5000}\)
Rõ ràng \(A< A'\)
SUY RA \(A^2< AA'=\frac{2}{50000}=\frac{1}{2500}=\left(\frac{1}{50}\right)^2\)
Nên \(A< \frac{1}{50}=0,02\)
Chúc bạn học tốt ( -_- )
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So sánh A = 1 + \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ ....+ \(\frac{1}{49^2}\)+\(\frac{1}{50^2}\) và B = 2
\(A=\frac{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}}{\frac{100}{1}+\frac{49}{2}+...+\frac{2}{49}+\frac{1}{50}}\)= ?