Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=4+2^2+2^3+...+2^{2005}\)
\(2A=4+2^2+2^3+...+2^{2006}\)
\(2A-A=\left(4+2^2+2^3+...+2^{2006}\right)-\left(4+2^2+2^3+...+2^{2005}\right)\)
\(A=4+2^2+2^3+...+2^{2006}-4-2^2-2^3-...-2^{2005}\)
\(A=2^{2006}\)
Vậy A là 1 luỹ thừa của cơ số 2
\(B=5+5^2+...+5^{2021}\)
\(5B=5^2+5^3+...+5^{2022}\)
\(5B-B=\left(5^2+5^3+...+5^{2022}\right)-\left(5+5^2+...+5^{2021}\right)\)
\(4B=5^{2022}-5\)
\(B=\frac{5^{2022}-5}{4}\)
\(B+8=\frac{5^{2022}-5}{4}+8\)
\(B+8=\frac{5^{2022}-5}{4}+\frac{32}{4}\)
\(B+8=\frac{5^{2022}-5+32}{4}\)
\(B+8=\frac{5^{2022}+27}{4}\)
=> B + 8 k thể là số b/ph của 1 số tn
\(A=2^2+2^2+2^3+...+2^{2005}\\ 2A=2^3+2^3+2^4+...+2^{2006}\\ 2A-A=\left(2^3+2^3+2^4+...+2^{2006}\right)-\left(2^2+2^2+2^3+...+2^{2005}\right)\\ A=2^{2006}\)
Chi tiết:
\(A=4+2^2+2^3+...+2^{2005}\\ 2A=4\cdot2+2^3+2^4+...+2^{2006}\\ 2A-A=\left(4\cdot2+2^3+2^4+...+2^{2006}\right)-\left(4+2^2+2^3+...+2^{2005}\right)\\ A=4\cdot2+2^{2006}-4-2^2=2^{2006}\left(Đpcm\right)\)
a, \(A=1+2+2^2+2^3+...+2^{100}\)
=> \(2A=2+2^2+2^3+2^4+...+2^{101}\)
=> \(A=2A-A=2^{101}-1\)
=> \(A+1=2^{101}\)
b, \(B=3+3^2+3^3+...+3^{2005}\)
\(3A=3^2+3^3+3^4+....+3^{2006}\)
=> \(2A=3A-A=3^{2006}-3\)
=> \(2A+3=3^{2006}\)là lũy thừa của 3
=> Đpcm
a) Ta có: \(A=1+2+2^2+2^3+.....+2^{100}\)
\(\Rightarrow2A=2+2^2+2^3+........+2^{101}\)
Lấy 2A-A ta có:
\(2A-A=\left(2+2^2+2^3+2^4+.....+2^{101}\right)\)\(-\left(1+2+2^2+2^3+.......+2^{100}\right)\)
\(\Rightarrow A=2^{101}-1\)
\(\Rightarrow A+1=2^{101}-1+1\)
\(\Rightarrow A+1=2^{101}\)
b) Ta có: \(B=3+3^2+3^3+.....+3^{2005}\)
\(\Rightarrow3B=3^2+3^3+3^4+.....+3^{2006}\)
\(\Rightarrow3B-B=\left(3^2+3^3+3^4+....+3^{2006}\right)\)\(-\left(3+3^2+3^3+......+3^{2005}\right)\)
\(\Rightarrow2B=3^{2006}-3\)
\(\Rightarrow2B+3=3^{2006}-3+3\)
\(\Rightarrow2B+3=3^{2006}\)
Vậy 2B+3 là lũy thừa của 3 ĐPCM
\(\Rightarrow2A=8+2^3+...+2^{2022}\\ \Rightarrow2A-A=8+2^3+...+2^{2022}-4-2^2-...-2^{2021}\\ \Rightarrow A=8+2^{2022}-4-2^2=8-4-4+2^{2022}=2^{2022}\left(đpcm\right)\)
\(A=2^2+2^2+2^3+...+2^{2021}=2^3+2^4+...+2^{2021}=2^{2022}\left(đpcm\right)\)
a) \(A=1+2+2^2+...+2^{80}\)
\(2A=2+2^2+2^3+...+2^{81}\)
\(2A-A=2+2^2+2^3+...+2^{81}-1-2-2^2-...-2^{80}\)
\(A=2^{81}-1\)
Nên A + 1 là:
\(A+1=2^{81}-1+1=2^{81}\)
b) \(B=1+3+3^2+...+3^{99}\)
\(3B=3+3^2+3^3+...+3^{100}\)
\(3B-B=3+3^2+3^3+...+3^{100}-1-3-3^2-...-3^{99}\)
\(2B=3^{100}-1\)
Nên 2B + 1 là:
\(2B+1=3^{100}-1+1=3^{100}\)
2)
a) \(2^x\cdot\left(1+2+2^2+...+2^{2015}\right)+1=2^{2016}\)
Gọi:
\(A=1+2+2^2+...+2^{2015}\)
\(2A=2+2^2+2^3+...+2^{2016}\)
\(A=2^{2016}-1\)
Ta có:
\(2^x\cdot\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x\cdot\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=\dfrac{2^{2016}-1}{2^{2016}-1}=1\)
\(\Rightarrow2^x=2^0\)
\(\Rightarrow x=0\)
b) \(8^x-1=1+2+2^2+...+2^{2015}\)
Gọi: \(B=1+2+2^2+...+2^{2015}\)
\(2B=2+2^2+2^3+...+2^{2016}\)
\(B=2^{2016}-1\)
Ta có:
\(8^x-1=2^{2016}-1\)
\(\Rightarrow\left(2^3\right)^x-1=2^{2016}-1\)
\(\Rightarrow2^{3x}-1=2^{2016}-1\)
\(\Rightarrow2^{3x}=2^{2016}\)
\(\Rightarrow3x=2016\)
\(\Rightarrow x=\dfrac{2016}{3}\)
\(\Rightarrow x=672\)
\(A=4+2^2+2^3+...+2^{2005}\)
\(2A=8+2^3+2^4+...+2^{2006}\)
\(2A-A=\left(8+2^3+2^4+...+2^{2006}\right)-\left(4+2^2+2^3+...+2^{2005}\right)\)
\(A=8+2^{2006}-\left(4+2^2\right)\)
\(A=2^{2006}\)
suy ra đpcm.
Cảm ơn bạn nha