a) Ta có: \(A=1+3+3^2+...+3^{99}+3^{100}\)

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

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

<=> \(2A=3^{101}-1\)

=> \(A=\frac{3^{101}-1}{2}\)

b) Ta có: \(B=1+4+4^2+...+4^{100}\)

=> \(4B=4+4^2+4^3+...+4^{101}\)

=> \(4B-B=\left(4+4^2+...+4^{101}\right)-\left(1+4+...+4^{100}\right)\)

<=> \(3B=4^{101}-1\)

=> \(B=\frac{4^{101}-1}{3}\)

*) \(A=2^2-2^4+2^6-2^8+....+2^{98}-2^{100}\)

\(\Leftrightarrow4A=2^4-2^6+2^8-2^{10}+....+2^{100}-2^{101}\)

\(\Leftrightarrow5A=2^2-2^{101}\)

\(\Leftrightarrow A=\frac{2^2-2^{101}}{5}\)

*) \(B=3-3^3+3^5-3^7+...+3^{79}-3^{99}\)

làm tương tự

Câu a:

A = 2 + 2^2+ 2^3 +..+ 2^100

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

2A - A = 2^2 + 2^3 + ..+ 2^101 - (2 + 2^2+ 2^3 +..+ 2^100)

A = 2^2 + 2^3 + ... + 2^101 - 2 - 2^2 - 2^3 -...-2^100

A = (2^101 - 2) + (2^2 - 2^2) + (2^3 -2^3) + ...+ (2^100 - 2^100)

A = 2^101 - 2 + 0 + 0 + ...+ 0

A = 2^101 - 2

Câu b:

B = 4+ 4^2 + 4^3 + ...+ 4^n

4B = 4^2 + 4^3 + ..+ 4^n+1

4B - B = 4^2 + 4^3 + ..+ 4^n+1-(4+ 4^2 + 4^3 + ...+ 4^n)

3B = 4^2 + 4^3 + ..+ 4^n+1 - 4 - 4^2 - 4^3 - ...- 4^n

3B = (4^n+1- 4) + (4^2 -4^2)+..+(4^n-4^n)

3B = 4^n+1- 4 + 0 + 0

3B = 4^n+1 - 4

B = (4^n+1 - 4)/3

\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)

\(2A=2+2^2+2^3+...+2^{51}\)

\(2A-A=A=2^{51}-2^0\)

\(B=5+5^2+5^3+...+5^{99}+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)

\(5B-B=4B=5^{101}-5\)

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

\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)

\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)

\(3C+C=4C=3^{2011}+3\)

\(C=\frac{3^{2011}+3}{4}\)

\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)

\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)

\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)

\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)

\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)

A=20+21+22+23+...++23+...+250250

2�=2+22+23+...+2512A=2+22+23+...+251

2�−�=�=251−202A−A=A=251−20

�=5+52+53+...+599+5100B=5+52+53+...+599+5100

5�=52+53+54+...+5100+51015B=52+53+54+...+5100+5101

<...

P= 2+2²+2³+...+2⁹⁹+2¹⁰⁰

2P=2²+2³+2⁴+....+2¹⁰⁰+2¹⁰¹

2P=2+2¹⁰¹

P=\(\frac{2+2^{101}}{2}\)

\(P=2+2^2+2^3+...+2^{99}\)\(+2^{100}\)

\(\Rightarrow2P=2^2+2^3+2^4+...+2^{100}\)\(+2^{101}\)

\(\Rightarrow2P-P=2^{101}-2\)

\(\Rightarrow P=2^{101}-2\)

Câu a:

A = 1+ 3 + 3^2+ ..+ 3^100

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

3A - A = 3 + 3^2+ 3^3 + ...+ 3^101 - (1+ 3 + 3^2+ ..+ 3^100)

2A = 3 + 3^2 + 3^3 + ...+ 3^100 - 1 - 3 - 3^2-..-3^100

2A = (3^101-1)+(3^2-3^2)+..+(3^100-3^100)

2A = 3^101 - 1 + 0 + 0+ ...+ 0

2A = 3^101 - 1

A = (3^101 - 1)/2

Câu b:

A = 1+ 4 + 4^2+ ..+ 4^100

4A = 4 + 4^2+ 4^3 + ...+ 4^101

4A - A = 4 + 4^2+ 4^3 + ...+ 4^101 - (1+ 4 + 4^2+ ..+ 4^100)

4A - A= =4 + 4^2 + 4^3 + ...+ 4^100 - 1 - 4 - 4^2-..-4^100

3A = (4^101-1)+(4^2-4^2)+..+(4^100-4^100)

3A = 4^101 - 1 + 0 + 0+ ...+ 0

3A = 4^101 - 1

A = (4^101 - 1)/3

\(2A=2^2+2^3+2^4+...+2^{100}+2^{101}\)

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

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

2A-A=A rồi mà kq A=2^201-2 thôi

bài A và B nè bạn!

A=1+3+3²+...+3¹⁰⁰

3A=3+3²+3³+...+3¹⁰¹

=>3A+1=1+3+3²+...+3¹⁰⁰+3¹⁰¹=A+3¹⁰¹

=>3A-A=3¹⁰¹-1

2A=3¹⁰¹-1

A=(3¹⁰¹-1)/2

B=1+4+4²+...+4⁵⁰

4B=4+4²+...+4⁵¹

4B+1=1+4+4²+...+4⁵⁰+4⁵¹=B+4⁵¹

=>4B-B=4⁵¹-1

3B=4⁵¹-1

B=(4⁵¹-1)/3

Đặt A = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹

2A = 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰

2A - A = (2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰) - (2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹)

A = 2¹⁰⁰ - 2