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

\(=120+3^5.\left(3^1+3^2+3^3+3^4\right)+....+3^{97}.\left(3^1+3^2+3^3+3^4\right)\)

\(=1.120+3^5.120+...+3^{97}.120\)

\(=\left(1+3^5+...+3^{97}\right).120\)

\(\Rightarrow S⋮120\)

Vậy ........

Ta có ;

S = 3 + 3 ^{2 +} 3 ³ + ........ + 3 ⁹⁹ + 3 ¹⁰⁰

    = ( 3 + 3 ² + 3 ³ + 3 ⁴ + 3 ⁵) + .... + ( 3 ⁹⁶ + 3 ⁹⁷ + 3 ⁹⁸ + 3 ⁹⁹ + 3 ¹⁰⁰ )

    = 3 ( 1 + 3 + 3 ² + 3 ³ + 3 ⁴ ) + .... + 3 ⁹⁶ . ( 1 + 3 + 3 ² + 3 ³ + 3 ⁴ ) 

    = 3 . 121 + .... + 3 ⁹⁶ . 121

    = 121 . ( 3 + .... + 3 ⁹⁶ ) chia hết cho 121 ( Do 121 chia hết cho 121 )

Vậy S = 3 + 3 ^{2 +} 3 ³ + ........ + 3 ⁹⁹ + 3 ¹⁰⁰ chia hết cho 121

B=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+......+(3^97+3^98+3^99+3^100)

B=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+.......+3^97(1+3+3^2+3^3)

B=3.40+3^5.40+......+3^97.40

B=40.3.(1+3+3^2+.......+3^98+3^99)

B=120.(1+3+3^2+.........+3^98+3^99)

Suy ra B chia hết cho 120

cho B=3+3^2+3^3+...+3^100.chứng minh rằng B chia hết cho 120

Ta có :

A=3+3^2+3^3+...+3^100

B=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+...+(3^97+3^98+3^99+3^100)

B=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+....+3^97(1+3+3^2+3^3)

B=3.40+3^5.40+....+3^97.40

B=40.(3+3^5+...+3^97)chia hết cho 40

Vì B có 25 số lũy thừa cơ số 3 nên M chia hết cho 3.

 Suy ra, B chia hết cho 40 và 3 tức là B chia hết cho 120 
vậy A chia hết cho 120

a^3-a-12a=a(a^2-1)-12a=a(a+1)(a-1)-12a              (1)

ta có a(a+1)(a-1) chia hết  cho 6

12 chia hết cho 6

nên (1) chia hết cho 6

suy ra a^3-13a chia hết cho 6

bai toan nay kho qua

\(E=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(3E=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(3E-E=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)

\(2E=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(6E=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(6E-2E=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)

\(4E=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)

\(4E=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)

\(4E=3-\frac{203}{3^{100}}< 3\)

\(\Rightarrow4E< 3\)

\(\Rightarrow E< \frac{3}{4}\left(đpcm\right)\)

Bài 1:

Ta có: \(3+3^2+3^3+...+3^{100}\)

\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=120+3^5\left(3+3^2+3^3+3^4\right)+....+3^{96}\left(3+3^2+3^3+3^4\right)\)

\(=120+3^5.120+...+3^{96}.120\)

\(=120.\left(1+3^5+.....+3^{96}\right)\)

\(\Rightarrow3+3^2+3^3+3^4+....+3^{100}\)chia hết cho 120 (vì có chứa thừa số 120)