\(A=1+2+2^2+2^3+....+2^{30}\)

\(2.A=2+2^2+2^3+2^4+...+2^{30}\)

\(2.A-A=\left(2+2^2+2^3+2^4+...+2^{31}\right)-\left(1+2+2^2+2^3+...+2^{30}\right)\)

\(A=2^{31}-1\)

\(\Rightarrow A+1=2^{31}-1+1\)

\(\Rightarrow A+1=2^{31}\)

2 mũ 31

A=1+2+2²+2³+2⁴+2⁵+2⁶+2⁷+2⁸+2⁹+2¹⁰.....+2³⁰

2.a=2+2²+2³+2⁴+2⁵+2⁶.....+2³⁰

2.A-A=

(2+2²+2³+2⁴+2⁵+2⁶+2⁷+2⁸+2⁹+2¹⁰+2¹¹+2¹²+2¹³.....2³¹)

_

(1+2+2²+2³.....+2³¹)

A

=2³¹-1

=>A+1=2³¹-1+1

=>A+1=2³¹

Chúc bạn học tốt nhé

A=1+2+2²+...+2³⁰

2A=2+2²+2³+...+2³¹

2A-A=(2+2²+2³+...+2³¹)-(1+2+2²+...+2³⁰)

A=2³¹-1

=>A+1=2³¹-1+1=2³¹

a=1+2+2^2+.......+2^30

2a=2(1+2+2^2+...+2^30)

2a=2+2^2+2^3+...2^31

2a-a=(2+2^2+2^3+...+2^31)-(1+2+2^2+2^30)

triệt tiêu ta có

a=2^31-1

a+1=2^31-1+1

a+1=2^31

thank you very much

\(A=1+3+3^2+...+3^{41}\)

\(3A=3+3^2+3^3+...+3^{42}\)

\(3A-A=3+3^2+...+3^{42}-1-3-...-3^{41}\)

\(2A=3^{42}-1\)

\(A=\dfrac{3^{42}-1}{2}\)

Ta có: \(2A+1\)

\(=2\cdot\dfrac{3^{42}-1}{2}+1\)

\(=3^{42}-1+1\)

\(=3^{42}\)

\(=\left(3^2\right)^{21}\)

\(=9^{21}\)

Ta có: \(A=2+2^2+2^3+...+2^{100}\)

\(2A=2^2+2^3+2^4+...+2^{101}\)

\(2A-A=2^{101}-2\)

Hay \(A=2^{101}-2\)

Vậy \(A=2^{101}-2\)

_Học tốt_

bạn trả lời quá chậm

tự làm nha 

a) \(\left(\frac{1}{16}\right)^{25}\div\left(\frac{1}{2}\right)^{30}=\left(\frac{1}{2^4}\right)^{25}\div\left(\frac{1}{2}\right)^{30}=\left[\left(\frac{1}{2}\right)^4\right]^{25}\div\left(\frac{1}{2}\right)^{30}=\left(\frac{1}{2}\right)^{4.25}\div\left(\frac{1}{2}\right)^{30}\)

\(=\left(\frac{1}{2}\right)^{100}\div\left(\frac{1}{2}\right)^{30}=\left(\frac{1}{2}\right)^{100-30}=\left(\frac{1}{2}\right)^{70}\)

b) \(584^{100}\div292^{100}=\left(584-292\right)^{100}=292^{100}\)

c) \(125^4\cdot16^3=\left(5^3\right)^4\cdot\left(2^4\right)^3=5^{3\cdot4}\cdot2^{4\cdot3}=5^{12}\cdot2^{12}=\left(5+2\right)^{12}=7^{12}\)

a: \(A=8^2\cdot32^4=2^6\cdot2^{20}=2^{26}\)

b: \(B=27^3\cdot9^4\cdot243=3^9\cdot3^8\cdot3^5=3^{22}\)

4 mũ 25