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26 tháng 8 2020

Bài làm:

Ta có: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\)

=> \(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\)

=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2017}}\right)\)

<=> \(2B=1-\frac{1}{3^{2017}}\)

=> \(B=\frac{1}{2}-\frac{1}{3^{2017}.2}< \frac{1}{2}\)

=> \(B< \frac{1}{2}\)

22 tháng 9 2019

ko hiểu

22 tháng 9 2019

\(3.M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}\)

=> \(3M-M=2M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{38}}-\frac{1}{3}-\frac{1}{3^2}-...-\frac{1}{3^{39}}\)

=> \(2M=1-\frac{1}{3^{39}}\)

=> \(M=\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)\)

do \(1-\frac{1}{3^{39}}< 1\)

=> \(\frac{1}{2}\left(1-\frac{1}{3^{39}}\right)< \frac{1}{2}.1=\frac{1}{2}\)

Vay \(M< \frac{1}{2}\)

Chuc bn hoc tot !

26 tháng 6 2016

Bấm máy tính là ra mak ^^

26 tháng 6 2016

cutecuteo mik cần lời giải cụ thể bạn ạ

20 tháng 8 2017

​3​​1​​−​4​​3​​−(−​5​​3​​)+​72​​1​​−​9​​2​​−​36​​1​​+​15​​1​​
=\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{72}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}=​3​​1​​−​4​​3​​+​5​​3​​+​72​​1​​−​9​​2​​−​36​​1​​+​15​​1​​
=\left(\frac{1}{3}-\frac{2}{9}\right)+\left(-\frac{3}{4}-\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)+\frac{1}{72}=(​3​​1​​−​9​​2​​)+(−​4​​3​​−​36​​1​​)+(​5​​3​​+​15​​1​​)+​72​​1​​
=\left(\frac{3}{9}-\frac{2}{9}\right)+\left(-\frac{27}{36}-\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)+\frac{1}{72}=(​9​​3​​−​9​​2​​)+(−​36​​27​​−​36​​1​​)+(​15​​9​​+​15​​1​​)+​72​​1​​
=\frac{1}{9}+\frac{-7}{9}+\frac{2}{3}+\frac{1}{72}=​9​​1​​+​9​​−7​​+​3​​2​​+​72​​1​​
=-\frac{2}{3}+\frac{2}{3}+\frac{1}{72}=−​3​​2​​+​3​​2​​+​72​​1​​
=0+\frac{1}{72}=\frac{1}{72}=0+​72​​1​​=​72​​1​​

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