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Hok tốt 

Sửa đề đi bạn , là 121 chứ , 120 chia ko được

Gọi tổng trên là S

S = 3¹ + 3² + 3³ + .. + 3²⁰¹⁶

S = (3¹ + 3² + 3³ + 3⁴ + 3⁵) + ... + (3²⁰¹² + 3²⁰¹³ + 3²⁰¹⁴ + 3²⁰¹⁵ + 3²⁰¹⁶)

S = 3¹.(1 + 3 + 3² + 3³ + 3⁴) + ... + 3²⁰¹².(1 + 3 + 3² + 3³ + 3⁴)

S = 3¹.121 + ... + 3²⁰¹².121

S = 121.(3¹ + ... + 3²⁰¹²)

Vì tích trên chứa 121 => S chia hết cho 121

3¹ + 3² + 3³ + 3⁴ + .................... + 3²⁰¹²

= (3¹ + 3² + 3³ + 3⁴) + (3⁵ + 3⁶ + 3⁷ + 3⁸) +............. + ( 3¹⁰⁰⁹ + 3¹⁰¹⁰ + 3¹⁰¹¹ + 3¹⁰¹²)

= (3 + 9 + 27 + 81) + 3⁴.(3 + 9 + 27 + 81) + ................. + 3¹⁰⁰⁸.(3 + 9 + 27 + 81)

= 120 + 3⁴.120 + ................. + 3¹⁰⁰⁸.120

= 120 . (1 + 3⁴ +3¹⁰⁰⁸)

đề nó thế rồi bạn k sửa đc đâu

\(3^1+3^2+3^3+3^4+3^5+...+3^{2012}\)

\(=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(=\left(3^1+3^2+3^3+3^4\right)+3^4\left(3^1+3^2+3^3+3^4\right)+...+3^{2008}\left(3^1+3^2+3^3+3^4\right)\)

\(=120\left(1+3^4+...+3^{2008}\right)⋮120\)

Gọi A = 3¹ + 3² + 3³ + 3⁴ + .............................. + 3²⁰¹²

A = (3¹ + 3² + 3³ + 3⁴) + (3⁵ + 3⁶ + 3⁷ + 3⁸) + ................ + ( 3²⁰⁰⁹ + 3²⁰¹⁰ + 3²⁰¹¹ + 3²⁰¹²)

A = (3 + 9 + 27 + 81) + 3⁴.(3 + 9 + 27 + 81) + ................. + 3²⁰⁰⁸.(3 + 9 + 27 + 81)

A = 120 + 3⁴ . 120 + ........................ + 3²⁰⁰⁸.120

A = 120.(1 + 3⁴ + .............. + 3²⁰⁰⁸)

chỉ như thế thôi à

cảm ơn bạn nhé

3¹+3²+.... +3²⁰¹²

=( 3¹+3² +3³+3⁴)+(3⁵+3⁶+3⁷+3⁸)....+(3²⁰⁰⁹+3²⁰¹⁰+3²⁰¹¹+3²⁰¹²)

=120+3¹(3+3¹+3²+3³)+....+3²⁰⁰⁸(3+3¹+3²+3³)

=120+3¹(120)+...3²⁰⁰⁸(120)

=120(1+3¹+...+3²⁰¹²)

Suy ra tong chia het cho 120

dung 100% luon bai nay minh hoc roi 

k minh nha

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Ta có: 3¹+3²+3³+3⁴+3⁵+...+3²⁰¹²

=(3^1+3^2+3^3+3^4+3^5)+...+(3^2008+3^2009+^3^2010+3^2011+3^2012)

=(3*1+3*3+3*3^2+3*3^3+3*3^4)+...+(3^2008*1+3^2008*3+3^2008*3^2+3^2008*3^3+3^2008*3^4)

=3*(1+3+3^2+3^3+3^4)+....+3^2008*(1+3+3^2+3^3+3^4)

=3*121+...+3^2008*121

=(3+3^6+...+3^2008)*121

Vì 121 chia 120 dư 1

Nên 3¹+3²+3³+3⁴+3⁵+...+3²⁰¹² chia hết cho 120

*là nhân nha bạn

Đặt S=\(3\)\(+\)\(3^2\)\(+\)\(3^3\)\(+\)...............\(+\)\(3^{2012}\)

\(\Rightarrow\)S=[\(3\)\(+\)\(3^2\)\(+\)\(3^3\)\(+\)]\(+\)........................\(+\)[\(3^{2009}\)\(+\)\(3^{2010}\)\(+\)\(3^{2011}\)\(+\)\(3^{2012}\)]

\(\Rightarrow\)S=120\(+\).......................\(+\)\(3^{2008}\)[\(3\)\(+\)\(3^2\)\(+\)\(3^3\)\(+\)\(3^4\)]

\(\Rightarrow\)S=120\(+\).......................\(+\)\(3^{2008}\)\(+\)120

\(\Rightarrow\)S=120[1\(+\)................\(+\)\(3^{2008}\)]

VÌ 120\(⋮\)120 \(\Rightarrow\)S\(⋮\)120

A=3+3²+3³+3⁴+.......+3¹⁰⁰

⇒A=(3+3²+3³+3⁴)+.......+(3⁹⁷+3⁹⁸+3⁹⁹+3¹⁰⁰)

⇒A=3.(1+3+3²+3³)+........+3⁹⁷.(1+3+3²+3³)

⇒A=3.40+.........+3⁹⁷.40

⇒A=40.(3+.......+3⁹⁷)

⇒A⋮40( 1 )

Vì A là tổng của các bậc lũy thừa của 3 nên A⋮3 ( 2 )

Từ ( 1 ) và ( 2 ) suy ra : A⋮ 40.3

⇒A⋮120

Vậy A⋮120( ĐPCM )

ta có 3^1+3^2+........+3^2012

=>(3^1+3^2+3^3+3^4)+.........+3^2009(3^1+3^2+3^3+3^4)

=>120+........................................+3^2009*120

=>120*(1+...............+3^2009) chia hết cho 120

vậy 3^1+3^2+.............+3^1012 chia hết cho 120

\(3+3^2+3^3+...+3^{2012}\)

\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=\left(3+3^2+3^3+3^4\right)+3.\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)

mà \(3+3^2+3^3+3^4=3+9+27+81=120⋮120\)

\(\Rightarrow\hept{\begin{cases}3+3^2+3^3+3^4⋮120\\3\left(3+3^2+3^3+3^4\right)⋮120\\3^{2008}\left(3+3^2+3^3+3^4\right)⋮120\end{cases}.......}\)

\(\Rightarrow3+3^2+3^3+...+3^{2012}⋮120\)

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