Bài 1:
$A=2^1+2^2+2^3+2^4$

$2A=2^2+2^3+2^4+2^5$

$\Rightarrow 2A-A=2^5-2^1$

$\Rightarrow A=2^5-1=32-1=31$

----------------------------

$B=3^1+3^2+3^3+3^4$

$3B=3^2+3^3+3^4+3^5$

$\Rightarrow 3B-B = 3^5-3$

$\Rightarrow 2B = 3^5-3\Rightarrow B = \frac{3^5-3}{2}$

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$C=5^1+5^2+5^3+5^4$

$5C=5^2+5^3+5^4+5^5$

$\Rightarrow 5C-C=5^5-5$

$\Rightarrow C=\frac{5^5-5}{4}$

Bài 2: Sai đề bạn nhé. Bạn xem lại.

a: Đặt \(A=2+2^2+\cdots+2^{60}\)

Ta có: \(A=2+2^2+\cdots+2^{60}\)

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

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

\(=3\left(2+2^3+\cdots+2^{59}\right)\) ⋮3

Ta có: \(A=2+2^2+\cdots+2^{60}\)

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

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

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

TA có: \(A=2+2^2+\cdots+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

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

\(=15\left(1+2^5+\cdots+2^{57}\right)\) ⋮15

b: Ta có: \(B=1+3+3^2+\cdots+3^{1991}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\cdots+\left(3^{1989}+3^{1990}+3^{1991}\right)\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+\cdots+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+\cdots+3^{1989}\right)\) ⋮13

c: Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{1997}+3^{1998}\right)\)

\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+\cdots+3^{1996}\left(3+3^2\right)\)

\(=12\left(1+3^2+\cdots+3^{1996}\right)\) ⋮12

Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{1996}+3^{1997}+3^{1998}\right)\)

\(=\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+\cdots+3^{1995}\left(3+3^2+3^3\right)\)

\(=39\left(1+3^3+\cdots+3^{1995}\right)\) ⋮39

câu hỏi tương tự

 cứ di chuột vào câu hỏi ế

a) \(\left(1+2+2^2+...+2^7\right)\)

\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^6+2^7\right)\)

\(=\left(1+2\right)+2^2.\left(1+2\right)+...+2^6.\left(1+2\right)\)

\(=3+2^2.3+...+2^6.3\)

\(=3.\left(1+2^2+...+2^6\right)⋮3\left(đpcm\right)\)

a) Đặt A = 1 + 2 + 2² + 2³ + ... + 2⁷

Ta có:

A = 1 + 2 + 2² + 2³ + ... + 2⁷

\(\Rightarrow\)2A = 2 + 2² + 2³ + 2⁴ + ... + 2⁸

\(\Rightarrow\)A = 2⁸ - 1 = 255

Vì 255\(⋮\)3\(\Rightarrow\)2 + 2² + 2³ + 2⁴ + ... + 2⁸\(⋮\)3

\(\Rightarrow\)ĐPCM

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{19}\right)⋮7\)

tự làm nha

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\left(2+...+2^{19}\right)⋮7\)

a: \(A=2\left(1+2+2^2\right)+...+2^{19}\left(1+2+2^2\right)\)

\(=7\cdot\left(2+...+2^{19}\right)⋮7\)