Giúp mình
\(1/2 + 1/4 + 1/6 + 1/8 + ... + 1/114 + 1/116\)
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9 tháng 10 2016
1 + 2 + 3 + 4 + 5 + 6 + .... + x = 55
suy ra (1 + x) * [(x - 1) + 1] / 2 = 55
hay (1 + x) * [(x - 1) + 1] = 55 * 2
hay (1 + x) * [(x - 1) + 1] = 110
Ta có: 110 = 11 * 10
suy ra (1 + x) ∈ { 10; 11} và [(x - 1) + 1] ∈ { 10; 11}
- Nếu 1 + x = 10
thì x = 10 - 1
hay x = 9
Thử lại: [(x - 1) + 1] = [(9 - 1) + 1] = 9 (loại vì 9 khác 11)
- Nếu 1 + x = 11
thì x = 11 - 1
hay x = 10
Thử lại: [(x - 1) + 1] = [(10 - 1) + 1] = 10 (nhận)
Vậy x = 10
18 tháng 1 2019
ta có: \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{x}.\)
\(A=1+\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{x}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2.2}+...+\frac{1}{x:2}\)
\(\Rightarrow2A-A=2-\frac{1}{x}\)
\(A=2-\frac{1}{x}=\frac{4095}{2048}\)
=> 1/x = 1/2048
=> x = 2048 ( 2048 = 211 )
9 tháng 8 2016
\(\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{9}.\frac{1}{10}\)
\(=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{10}\)
\(=\frac{9}{10}\)
PN
1
15 tháng 8 2016
Đặt A= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
=\(\frac{1}{2^1}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
=> 2A= \(1-\frac{1}{2^1}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
Ta có:
2A+A=\(\left(1-\frac{1}{2^1}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2^1}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
=> 3A=\(1-\frac{1}{2^1}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}+\frac{1}{2^1}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
=\(1-\left(\frac{1}{2^1}-\frac{1}{2^1}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)-\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)-\left(\frac{1}{2^5}-\frac{1}{2^5}\right)-\frac{1}{2^6}\)
= \(1-\frac{1}{2^6}\)
=> A= 3A:3= \(\left(1-\frac{1}{2^6}\right):3\)=\(\frac{1}{3}-\frac{1}{2^6}:3\)<\(\frac{1}{3}\)
18 tháng 1 2019
\(2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{2}{x}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{2}{x}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{2}{x}+\frac{1}{x}\right)\)
=> \(A=2-\frac{1}{x}\)
Giải phương trình:
\(2-\frac{1}{x}=\frac{4095}{2048}\)
\(\frac{1}{x}=2-\frac{4095}{2048}\)
\(\frac{1}{x}=\frac{1}{2048}\)
x=2048
Ta có:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{114}+\frac{1}{116}\)
\(=\frac{1}{1\text{x}2}+\frac{1}{2\text{x}2}+\frac{1}{2\text{x}3}+\frac{1}{2\text{x}4}+...+\frac{1}{57\text{x}2}+\frac{1}{58\text{x}2}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{57}-\frac{1}{2}+\frac{1}{58}-\frac{1}{2}\)
\(=1-\frac{1}{2}-\frac{1}{3}-\frac{1}{2\text{x}2}-\frac{1}{2\text{x}2\text{x}2}-...-\frac{1}{57}-\frac{1}{29\text{x}2}\)
\(=1-\frac{1}{2}-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}-...-\frac{1}{57}-\frac{1}{29}-\frac{1}{2}\)
\(=1-\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-...-\frac{1}{29}\)
\(=1-\frac{1}{29}\)
\(=\frac{28}{29}\)