a: \(B=3\left(1+3+3^2+...+3^{120}\right)⋮3\)

b: \(B=4\left(3+...+3^{119}\right)⋮4\)

câu hỏi tương tự

 cứ di chuột vào câu hỏi ế

Bài 1:
$A=2^1+2^2+2^3+2^4$

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

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

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

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

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

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

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

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

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

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

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

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

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

Bài 2: Sai đề bạn nhé. Bạn xem lại.

* ta có : \(C=3^1+3^2+3^3+...+3^{99}+3^{100}\) có \(100\) số hạng

và \(100⋮4\) và \(100⋮̸3\)

ta có : \(C=3^1+3^2+3^3+...+3^{99}+3^{100}\)

\(=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\) (vì \(100⋮4\) )

\(=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+2^{97}\left(1+3+3^2+3^3\right)\)

\(=3\left(1+3+9+27\right)+3^5\left(1+3+9+27\right)+...+2^{97}\left(1+3+9+27\right)\)

\(=3.40+3^5.40+...+3^{97}.40=40.\left(3+3^5+...+3^{97}\right)⋮40;10;4\)

vậy \(C\) chia hết cho \(40;10và4\) (1)

ta có : \(C=3^1+3^2+3^3+...+3^{99}+3^{100}\)

\(=3^1+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\) (vì \(100⋮̸3\) )

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

\(=3+3^2\left(1+3+9\right)+3^5\left(1+3+9\right)+...+2^{98}\left(1+3+9\right)\)

\(=3+3^2.13+3^5.13+...+3^{98}.13=3+13.\left(3^2+3^5+...+3^{98}\right)\)

ta có : \(13.\left(3^2+3^5+...+3^{98}\right)⋮13\) nhưng \(3⋮̸13\)

\(\Rightarrow\) \(C\) không chia hết cho \(13\) và \(3< 13\) \(\Rightarrow\) \(3\) là số dư khi chia \(C\) cho \(13\) (2)

từ (1) và (2) \(\Rightarrow\) (ĐPCM)

a: \(D=3^2+3^4+...+3^{120}\)

\(=3\cdot3+3\cdot3^3+...+3\cdot3^{119}\)

\(=3\left(3+3^3+...+3^{119}\right)⋮3\)

b: \(D=3^2+3^4+3^6+...+3^{120}\)

\(=3^2+3^2\cdot3^2+3^2\cdot3^4+...+3^2\cdot3^{118}\)

\(=3^2\left(1+3^2+3^4+...+3^{114}+3^{116}+3^{118}\right)\)

\(=9\cdot\left[\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{114}\left(1+3^2+3^4\right)\right]\)

\(=9\cdot91\left[1+3^6+...+3^{114}\right]⋮91\)

Anh giải cho em câu em mới đăng với ạ

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