2A = 3A - A = (3 + 3² + 3³ + ... + 3¹⁰¹) - (1 + 3 + 3² + 3³ + ... + 3¹⁰⁰)

2A = 3¹⁰¹ - 1

A = \(\frac{3^{101}-1}{2}\)

3B = 4B - B = (4 + 4² + ... + 4⁵¹) - (1 + 4 + 4² + ... + 4⁵⁰)

3B = 4⁵¹ - 1

B = \(\frac{4^{51}-1}{3}\)

Đề bài là gì zợ !

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

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2B-B=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)

\(\Rightarrow2B-B=2^{101}-2\)

bài 1

2¹⁰¹ - 2

a: A=2/9(9+99+...+99..99)

=2/9(10-1+10^2-1+...+10^22-1)

=2/9[10+10^2+...+10^22-22]

Đặt B=10+10^2+...+10^22

=>10B=10^2+10^3+...+10^23

=>B=(10^23-10)/9

=>\(A=\dfrac{2}{9}\cdot\left(\dfrac{10^{23}-10}{9}-22\right)\)

=>\(A=\dfrac{2\cdot10^{23}-416}{81}\)

\(S=\left(1+4\right)+\left(4^2+4^3\right)+...+\left(4^{98}+4^{99}\right)\\ S=\left(1+4\right)+4^2\left(1+4\right)+...+4^{98}\left(1+4\right)\\ S=\left(1+4\right)\left(1+4^2+...+4^{98}\right)=5\left(1+4^2+...+4^{98}\right)⋮5\)

\(S=\left(1+4\right)+...+4^{98}\left(1+4\right)\)

\(=5\left(1+...+4^{98}\right)⋮5\)

ta có 

\(1+3+3^2+..+3^{2000}=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+..+\left(3^{1998}+3^{1999}+3^{2000}\right)\)

\(=13.1+13\cdot3^3+..+13\cdot3^{1998}\) chia hết cho 13

tương tự

\(1+4+4^2+..+4^{2012}=\left(1+4+4^2\right)+..+\left(4^{2010}+4^{2011}+4^{2012}\right)\)

\(=21.1+21\cdot4^3+..+21.4^{2010}\) chia hết cho 21