Ta có:9¹¹+1=(9²)⁵.9+1=81⁵.9+1

=.........1.9+1

=...........9+1

=.............0 chia hết cho 2 và 5

\(A=11+13+15+....+99.\)

\(=\frac{\left(11+99\right)\left[\left(99-11\right):2+1\right]}{2}\)

\(=\frac{110.\left(88:2+1\right)}{2}\)

\(=\frac{110.45}{2}\)

\(=\frac{4950}{2}=2475\)

\(\Rightarrow\hept{\begin{cases}A⋮̸⋮2\\A⋮5\end{cases}}\)

(1+2³)+(2+2⁴)+...+(2⁸+2¹¹)

9+2(1+2³)+...+2⁸(1+2³)

9(1+2+...+2⁸) chia hết cho 9

=>( 2^0+2^1+2^2 + ...+2^11) chia hết cho 9

c)(5+5²)+(5³+5⁴)+...+(5⁹⁹+5¹⁰⁰)

5(1+5)+5³(1+5)+...+5⁹⁹(1+5)

6(5+5³+...+5⁹⁹) chia hết cho 3

Sửa đề: \(A=2^0+2^1+2^2+...+2^{99}\)

\(=\left(2^0+2^1\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)

\(=\left(1+2\right)+2^2\left(1+2\right)+...+2^{98}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{98}\right)⋮3\)

TA CÓ:

A=3⁰+3+3²+3³+........+3¹¹

(3⁰+3+3²+3³)+....+(3⁸+3⁹+3¹⁰+3¹¹)

3(0+1+3+3²)+......+3⁸(0+1+3+3²) 

3.13+....+3⁸.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)

\(A=1+2+2^2+2^3+...+2^{99}\)

\(=\left(1+2+2^2+2^3\right)+\left(2^4+2^5+2^6+2^7\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}\right)\)

\(=\left(1+2+2^2+2^3\right)+2^4\left(1+2+2^2+2^3\right)+...+2^{96}\left(1+2+2^2+2^3\right)\)

\(=15\left(1+2^4+...+2^{96}\right)⋮15\)

Giúp tớ với tớ cần gấp