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20 tháng 7 2015

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\)

\(\Rightarrow2A=2.\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)

\(=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}\)

\(\Rightarrow2A-A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2006}}\right)\)

\(\Rightarrow A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2005}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2006}}\)

\(=2-\frac{1}{2^{2006}}\)

4 tháng 8 2015

Tớ cx dg lm cái này :v

 

Ta có: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2}\)

\(=\dfrac{3^{32}-1}{2}\)

Rút gọn: (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

A=2(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)

A=(3-1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(32-1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(34-1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)
A=(38-1)(38 + 1)(316 + 1)(332 + 1)
A=(316-1)(316 + 1)(332 + 1)
A=(332 - 1)(332 + 1)
A=364-1
=>A=(364-1) /2

DD
28 tháng 3 2021

\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+111\right)}\)

\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{\left(1+1+...+1\right)+\left(2+2+...+2\right)+...+111}\)(\(111\)số hạng \(1\)\(110\)số hạng \(2\),...)

\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{1\times111+2\times110+3\times109+...+111\times1}\)

\(A=1\)

4 tháng 8 2018

B = (1 + 1/2)(1 + 1/3)(1 + 1/4) ...(1 + 1/100)

    = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{101}{100}\)

    = \(\frac{3.4.5....101}{2.3.4...100}=\frac{101}{2}\)

C = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{1000}\right)\)

   \(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{999}{1000}\)

    \(=\frac{1.2.3...999}{2.3.4....1000}=\frac{1}{1000}\)

26 tháng 7 2016

Tong quat: a^3+1=(a+1)[a^2-a+1]=(a+1)[(a-0,5)^2+0,75]

                 a^3-1=(a-1)[a^2+a+1]=(a-1)[(a+0,5)^2+0,75]

Tu so cua A=(2+1).[(2-0,5)^2+0,75].(3+1).[(3-0,5)^2+0,75].(4+1).[(4-0,75)^2+0,75]....(10+1).[(10-0,5)^2+0,75]

                 =3.[1,5^2+0,75].4.[2,5^2+0,75].5.[3,5^2+0,75]....11.[9,5^2+0,75]

Mau so cua A= (2-1).[(2+0,5)^2+0,75].(3-1).[(3+0,5)^2+0,75].(4-1).[(4+0,75)^2+0,75]....(10-1).[(10+0,5)^2+0,75]

                 =[2,5^2+0,75].2.[3,5^2+0,75].3.[4,5^2+0,75]....9.[10,5^2+0,75]

Vay A=3.[1,5^2+0,75].4.[2,5^2+0,75].5.[3,5^2+0,75]....11.[9,5^2+0,75]/[2,5^2+0,75].2.[3,5^2+0,75].3.[4,5^2+0,75]....9.[10,5^2+0,75]

         =(3.4.5...11/1.2.3...9).[(1,5^2+0,75)(2,5^2+0,75)(3,5^2+0,75)...(9,5^2+0,75)/(2,5^2+0,75)(3,5^2+0,75)(4,5^2+0,75)...(10,5^2+0,75)]

         =11.10.(1,5^2+0,75)/2.(10,5^2+0,75)

         Con bao nhieu ban tu tinh tiep nha 

Tai vi may minh bi lag nen khong danh phan so duoc vi vay minh phai tach mau, tu ra. sorry

26 tháng 7 2016

cảm ơn bạn nhiều