\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)

=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)

=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

=>\(A=1-\frac{1}{2^{100}}\)

1/2.A=1/2²+1/2³+...+1/2¹⁰¹

=>1/2A-A=1/2¹⁰¹-1/2

=>-1/2A=1/2¹⁰¹-1/2

A=(1/2¹⁰¹-1/2):(-1/2)=(1/2¹⁰¹-1/2).(-2)

=1-1/2¹⁰⁰

\(A=\dfrac{17^{100}+17^{96}+17^{92}+....+17^4+1}{17^{102}+17^{100}+17^{98}+....+17^2+1}\)

Gọi \(17^{100}+17^{96}+17^{92}+....+17^4+1\) là B

\(B=17^{100}+17^{96}+17^{92}+....+17^4+1\\ 17^4\cdot B=17^{104}+17^{100}+17^{96}+......+17^8+17^4\\ 17^4\cdot B-B=\left(17^{104}+17^{100}+17^{96}+......+17^8+17^4\right)-\left(17^{100}+17^{96}+17^{92}+....+17^4+1\right)\\ B\cdot\left(17^4-1\right)=17^{104}-1\\ B=\dfrac{17^{104}-1}{17^4-1}\)

Gọi \(17^{102}+17^{100}+17^{98}+....+17^2+1\) là C

\(C=17^{102}+17^{100}+17^{98}+....+17^2+1\\ C\cdot17^2=17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\\ C\cdot17^2-C=\left(17^{104}+17^{102}+17^{100}+17^{98}+....+17^2\right)-\left(17^{102}+17^{100}+17^{98}+....+17^2+1\right)\\ C\cdot\left(17^2-1\right)=17^{104}-1\\ C=\dfrac{17^{104}-1}{17^2-1}\)

=>

 \(A=B:C\\ A=\dfrac{17^{104}-1}{17^4-1}:\dfrac{17^{104}-1}{17^2-1}\\ A=\dfrac{17^2-1}{17^4-1}\)

cảm ơn bạn

A = \(\dfrac{T}{M}\)

M = T + (\(17^{102}+17^{98}+17^{94}+...+17^2\))

M = T + \(17^2\left(17^{100}+17^{96}+17^{92}+...+17^0\right)\)

M = T + \(17^2\cdot\) T = T(\(1+17^2\))

=> A = \(\dfrac{T}{T\left(1+17^2\right)}=\dfrac{1}{1+17^2}\)

Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)

\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}=1-\dfrac{1}{2^{100}}\)

\(A=1+5+5^2+5^3+...+5^{100}\)

\(5A=5+5^2+5^3+...+5^{100}+5^{101}\)

\(5A-A=5+5^2+5^3+...+5^{100}+5^{101}-\left(1+5+5^2+5^3+...+5^{100}\right)\)

\(4A=5^{101}-1\)

\(A=\frac{5^{101}-1}{4}\)

A = 1+5+5²+5³+...+5¹⁰⁰

5A = 5+5²+5³+5⁴+....+5¹⁰¹

4A = 5A - A = 5¹⁰¹ - 1

=> A = \(\frac{5^{101}-1}{4}\)

thank you

A=1+2^2+...+2^100

2A=2+2^2+2^3+...+2^101

2A=2^101-1

A=(2^101-1):2

\(B=5^1+5^2+...+5^{199}\)

\(\Rightarrow5B=5^2+5^3+...+5^{200}\)

\(\Rightarrow5B-B=\left(5^2+5^3+...+5^{200}\right)-\left(5^1+5^2+...+5^{199}\right)\)

\(\Rightarrow4B=5^{200}-5\)

\(\Rightarrow B=\frac{5^{200}-5}{4}\)

Bài 1: 

a: \(2A=2^{101}+2^{100}+...+2^2+2\)

\(\Leftrightarrow A=2^{100}-1\)

b: \(3B=3^{101}+3^{100}+...+3^2+3\)

\(\Leftrightarrow2B=3^{100}-1\)

hay \(B=\dfrac{3^{100}-1}{2}\)

c: \(4C=4^{101}+4^{100}+...+4^2+4\)

\(\Leftrightarrow3C=4^{101}-1\)

hay \(C=\dfrac{4^{101}-1}{3}\)

Trả lời

A = 1 + 2¹ + 2² + ... + 2⁹⁹ + 2¹⁰⁰ 

2A = 2 + 2² + 2³ + ... + 2¹⁰⁰ + 2¹⁰¹

2A - A = A = ( 2 + 2² + 2³ + ... + 2¹⁰⁰ + 2¹⁰¹ ) - ( 1 + 2¹ + 2² + ... + 2⁹⁹ + 2¹⁰⁰ )

A = 2¹⁰¹ - 1

\(A=1+2^1+2^2+...+2^{99}+2^{100}\)

\(2A=2+2^2+...+2^{100}+2^{101}\)

Ta có:\(2A-A=\left(2^1+2^2+...+2^{100}\right)-\left(1+2^1+2^2+...+2^{101}\right)\)

\(A=2^{101}-1\)

#hok tốt#

Trả lời :

A = 2¹⁰⁰ - 1

~ Hok tốt ~