a) A = 1 + 4 + 4² + 4³ + ... + 4¹¹

\( = \left( {1 + 4 + {4^2}} \right) + \left( {{4^3} + {4^4} + {4^5}} \right) + ... + \left( {{4^9} + {4^{10}} + {4^{11}}} \right)\)

\( = \left( {1 + 4 + {4^2}} \right) + {4^3}\,.\,\left( {1 + 4 + {4^2}} \right) + ... + {4^9}\,.\,\left( {1 + 4 + {4^2}} \right)\)

\( = \left( {1 + 4 + {4^2}} \right)\,.\,\left( {1 + {4^3} + ... + {4^9}} \right)\)

\( = 21\,.\,\left( {1 + {4^3} + ... + {4^9}} \right)\; \vdots \;21\)

Vậy A ⋮ 21.

b) A = 1 + 4 + 4² + 4³ + ... + 4¹¹

\( = \left( {1 + 4} \right) + \left( {{4^2} + {4^3}} \right) + ... + \left( {{4^{10}} + {4^{11}}} \right)\)

\( = \left( {1 + 4} \right) + {4^2}\,.\,\left( {1 + 4} \right) + ... + {4^{10}}\,.\,\left( {1 + 4} \right)\)

\( = \left( {1 + 4} \right)\,.\,\left( {1 + {4^2} + ... + {4^{10}}} \right)\)

\( = 5\,.\,\left( {1 + {4^2} + ... + {4^{10}}} \right)\; \vdots \;5\)

Vậy A ⋮ 5

Với A ⋮ 5 và A ⋮ 21 mà ƯCLN(5; 21) = 1 nên A ⋮ 5 × 21 hay A ⋮ 105.

c) A = 1 + 4 + 4² + 4³ + ... + 4¹¹

\( = \left( {1 + {4^2}} \right) + \left( {4 + {4^3}} \right) + ... + \left( {{4^8} + {4^{10}}} \right) + \left( {{4^9} + {4^{11}}} \right)\)

\( = \left( {1 + {4^2}} \right) + 4\,.\,\left( {1 + {4^2}} \right) + ... + {4^8}\,.\,\left( {1 + {4^2}} \right) + {4^9}\left( {1 + {4^2}} \right)\)

\( = \left( {1 + {4^2}} \right)\,.\,\left( {1 + 4 + {4^4} + {4^5} + {4^8} + {4^9}} \right)\)

\( = 17\,.\,\left( {1 + 4 + {4^4} + {4^5} + {4^8} + {4^9}} \right)\; \vdots \;17\)

Vậy A ⋮ 17.

Xét \(B = 1 + 4 + {4^4} + {4^5} + {4^8} + {4^9}\)

\( = \left( {1 + {4^4} + {4^8}} \right) + \left( {4 + {4^5} + {4^9}} \right)\)

\( = \left( {1 + {4^4} + {4^8}} \right) + 4\left( {1 + {4^4} + {4^8}} \right)\)

\( = 5\,.\,\left( {1 + {4^4} + {4^8}} \right) = 5\,\,.\,\,65\,\,793\)

Vì 65793 ⋮ 241 nên B ⋮ 241 suy ra A ⋮ 241.

Với A ⋮ 17 và A ⋮ 241 mà ƯCLN(17; 241) = 1 nên A ⋮ 17 × 241 hay A ⋮ 4097.

Chứng minh bất đẳng thức: a^2 + b^2 + c^2 > = ab + bc + ca