Question

[bao123]

Cho A = 2 + 2 mũ 2 + 2 mũ 3 + .... + 2 mũ 60 a ) Thu gọn tổng A b) Chứng minh rằng : A chia hết cho 3,5, 7

Answer 1

[Toru]

a) \(A=2+2^2+2^3+\dots+2^{60}\)

\(2A=2^2+2^3+2^4+\dots+2^{61}\)

\(2A-A=\left(2^2+2^3+2^4+\dots+2^{61}\right)-\left(2+2^2+2^3+\dots+2^{60}\right)\)

\(A=2^{61}-2\)

Vậy: \(A=2^{61}-2\).

b)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\dots+\left(2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+2^5\cdot\left(1+2\right)+\dots+2^{59}\cdot\left(1+2\right)\)

\(=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{59}\cdot3\)

\(=3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)\)

Vì \(3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)⋮3\) nên \(A⋮3\)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\dots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+2^9\cdot\left(1+2+2^2+2^3\right)+\dots+2^{57}\cdot\left(1+2+2^2+2^3\right)\)

\(=2\cdot15+2^5\cdot15+2^9\cdot15+\dots+2^{57}\cdot15\)

\(=15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)\)

Vì \(15⋮5\) nên \(15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)⋮5\)

hay \(A\vdots5\)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\dots+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+2^7\cdot\left(1+2+2^2\right)+\dots+2^{58}\cdot\left(1+2+2^2\right)\)

\(=2\cdot7+2^4\cdot7+2^7\cdot7+\dots+2^{58}\cdot7\)

\(=7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)\)

Vì \(7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)⋮7\) nên \(A⋮7\)

$Toru$

Answer 2

[LÊ DƯƠNG QUỲNH TRÂM]

a) $A=2+2^2+2^3+\cdots +2^{60}$

$2A=2^2+2^3+2^4+\cdots +2^{61}$

$2A-A=\left(2^2+2^3+2^4+\cdots +2^{61}\right)-\left(2+2^2+2^3+\cdots +2^{60}\right)$

$A=2^{61}-2$

Vậy: $A=2^{61}-2$.

b)

+) $A=2+2^2+2^3+\cdots +2^{60}$

$=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\cdots +\left(2^{59}+2^{60}\right)$

$=2\cdot \left(1+2\right)+2^3\cdot \left(1+2\right)+2^5\cdot \left(1+2\right)+\cdots +2^{59}\cdot \left(1+2\right)$

$=2\cdot 3+2^3\cdot 3+2^5\cdot 3+\cdots +2^{59}\cdot 3$

$=3\cdot \left(2+2^3+2^5+\cdots +2^{59}\right)$

Vì $3\cdot \left(2+2^3+2^5+\cdots +2^{59}\right)⋮3$ nên $A⋮3$

+) $A=2+2^2+2^3+\cdots +2^{60}$

$=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\cdots +\left(2^{57}+2^{58}+2^{59}+2^{60}\right)$

$=2\cdot \left(1+2+2^2+2^3\right)+2^5\cdot \left(1+2+2^2+2^3\right)+2^9\cdot \left(1+2+2^2+2^3\right)+\cdots +2^{57}\cdot \left(1+2+2^2+2^3\right)$

$=2\cdot 15+2^5\cdot 15+2^9\cdot 15+\cdots +2^{57}\cdot 15$

$=15\cdot \left(2+2^5+2^9+\cdots +2^{57}\right)$

Vì $15⋮5$ nên $15\cdot \left(2+2^5+2^9+\cdots +2^{57}\right)⋮5$

hay $A⋮5$

+) $A=2+2^2+2^3+\cdots +2^{60}$

$=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\cdots +\left(2^{58}+2^{59}+2^{60}\right)$

$=2\cdot \left(1+2+2^2\right)+2^4\cdot \left(1+2+2^2\right)+2^7\cdot \left(1+2+2^2\right)+\cdots +2^{58}\cdot \left(1+2+2^2\right)$

$=2\cdot 7+2^4\cdot 7+2^7\cdot 7+\cdots +2^{58}\cdot 7$

$=7\cdot \left(2+2^4+2^7+\cdots +2^{58}\right)$

Vì $7\cdot \left(2+2^4+2^7+\cdots +2^{58}\right)⋮7$ nên $A⋮7$