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![](https://rs.olm.vn/images/avt/0.png?1311)
S = (1 - 3 + 32 - 33) + 34 . (1 - 3 + 32 - 33) + .... + 396 . (1 - 3 + 32 - 33)
S = (-20) + 34 . (-20) +.... + 396 . (-20)
S = (-20) . (1 + 34 +...+ 396)
\(\Rightarrow\)S \(⋮\) 20
(Ko bt có đúng ko)
*KO CHÉP MẠNG*
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1-3+3^2-3^3+...+3^{98}-3^{99}=\left(1-3+3^2-3^3\right)+3^4\left(1-3+3^3-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)=\left(-20\right)+3^4.\left(-20\right)+...+3^{96}.\left(-20\right)=\left(-20\right)\left(1+3^4+...+3^{96}\right)⋮20\)
Ta có: \(S=1-3+3^2-3^3+...+3^{98}-3^{99}\)
\(=\left(1-3+3^2-3^3\right)+...+3^{96}\left(1-3+3^2-3^3\right)\)
\(=-20\cdot\left(1+...+3^{96}\right)⋮20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)
\(=13+3^3.13+...+3^{96}.13=13\left(1+3^3+...+3^{96}\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
S = 1 - 3 + 32 - 33+...+398 - 399
S = 30 - 31 + 32 - 33+...+ 398 - 399
xét dãy số: 0; 1; 2; 3;...;99
Dãy số trên là dãy số cách đều với khoảng cách là: 1 - 0 = 1
Dãy số trên có số số hạng là: (99 - 0): 1 + 1 = 100 (số)
100 : 4 = 25
Vậy ta nhóm 4 số hạng liên tiếp của tổng S thành 1 nhóm thì:
S = ( 1 - 3 + 32 - 33) +....+( 396 - 397 + 398 - 399)
S = - 20+...+ 396.(1 - 3 + 32 - 33)
S = - 20 +...+ 396.(-20)
S = -20.( 30 + ...+ 396) (đpcm)
b,
S = 1 - 3 + 32 - 33+...+ 398 - 399
3S = 3 - 32 + 33-...-398 + 399 - 3100
3S + S = - 3100 + 1
4S = - 3100 + 1
S = ( -3100 + 1): 4
S = - ( 3100 - 1) : 4
Vì S là số nguyên nên 3100 - 1 ⋮ 4 ⇒ 3100 : 4 dư 1 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107.101107`
\(A = 1 + 3 + 3^2 + 3^3 + ... + 3^{98} + 3^{99}\)
\(A = (1 + 3) + (3^2 + 3^3) + ... + (3^{98} + 3^{99})\)
\(A = (1 + 3) + 3^2(1 + 3) + ... + 3^{98}(1 + 3)\)
\(A = (1 + 3)(1 + 3^2 + ... + 3^{98})\)
\(A = 4(1 + 3^2 + ... + 3^{98})\)
Vì \(4(1 + 3^2 + ... + 3^{98}) \) \(\vdots\) \(4\)
`\Rightarrow A \vdots 4`
Vậy, `A \vdots 4` (đpcm).
A = 1 + 3 + 32 + 33 + ... + 398 + 399
A = (1 + 3) + (32 + 33) + ... + (398 + 399)
A = 1. (1 + 3) + 32. (1 + 3) + ... + 398. (1 + 3)
A = 1.4 + 32.4 + ... + 398.4
A = 4. (1 + 32 + ... + 398)
⇒ A ⋮ 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)
\(=13+3^3.13+...+3^{96}.13\)
\(=13\left(1+3^3+...+3^{96}\right)⋮13\)
\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ A=13\left(1+3^3+...+3^{96}\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a. $2^{29}< 5^{29}< 5^{39}$
$\Rightarrow A< B$
b.
$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$
$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$
$=(1+3)(3+3^3+3^5+...+3^{2009})$
$=4(3+3^3+3^5+...+3^{2009})\vdots 4$
Mặt khác:
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$
$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$
Bài 1:
c.
$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$
$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$
$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$
$\Rightarrow A=\frac{3^{101}+1}{4}$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
S=1-3+32-...+398-399
=(1-3+32-33)+...+(396-397+398-399)
=-20+...+396.(-20)
=-20.(1+....+396)
nên S chhia hết cho(-20) (đpcm)
=>(1-3+32-33)+...+(396-397+398-399)
=> -20+...+396.(-20)
=>-20.(1+....+396)
Nên S chhia hết cho(-20) (đpcm)
tick nhé