S=4+32+33+...+3223

S=1+3+32+33+...+3223

S=(1+34)+(3+35)+(32+36)+(33+37)+...+(3119+3223)

S=82+3(1+34)+32(1+34)+33(1+34)+...+3119(1+34)

S=82+3.82+32.82+33.82+...+3119.(1+34)

S=82(3+32+33+...+3119)

vì 82⋮41⇒S⋮41

Vậy S⋮41

Ta co: 3+3^3+3^5+...+3^1991 = (3+3^3+3^5)+...+(3^1987+1989+1991) =3.(1+3^2+3^4)+...+3^1987.(1+3^2+3^4) =3.91+...+3^1987.91 =(3+..+3^1987).91=(3+...+3^1987).13.7 chia het cho 13 3+3^3+3^5+...+3^1991 =(3+3^3+3^5+3^7)+...+(3^1985+3^1987+3^1989+3^1991) =3(1+3^2+3^4+3^6)+...+3^1985.(1+3^2+3^4+3^6) =3.820+...+3^1985.820=(3+...+3^1985).820=(3+....+3^1985).41.20 chia het cho 41

WHY CHO 3^223 CƠ MÀ

S=4+3²+3³+...+3²²³

S=1+3+3²+3³+...+3²²³

S=(1+3⁴)+(3+3⁵)+(3²+3⁶)+(3³+3⁷)+...+(3¹¹⁹+3²²³)

S=82+3(1+3⁴)+3²(1+3⁴)+3³(1+3⁴)+...+3¹¹⁹(1+3⁴)

S=82+3.82+3².82+3³.82+...+3¹¹⁹.(1+3⁴)

S=82(3+3²+3³+...+3¹¹⁹)

vì 82⋮41⇒S⋮41

Vậy S⋮41

\(S=4+3^2+3^3+...+3^{223}=3^0+3^1+3^2+3^3+...+3^{223}\)

=> \(3S=3+3^2+3^3+3^4+...+3^{224}\)

=> \(3S-S=3^{224}-1\)

=> \(S=\frac{3^{224}-1}{2}=\frac{\left(3^8\right)^{28}-1}{2}\)là số tự nhiên 

Ta có: \(\left(3^8\right)^{28}-1⋮\left(3^8-1\right)\)

mà \(3^8-1=6560=41.160⋮41\) 

=> \(\left(3^8\right)^{28}-1⋮41;\left(41;2\right)=1\)

=> \(S=\frac{\left(3^8\right)^{28}-1}{2}\) chia hết cho 41.

Thank nha !

😊😊😊😊

Đề sai nha

S=3+3²+3³+...+3²²³

S=(3+3²+3³+3⁴+3⁵+3⁶+3⁷+3⁸)+.....+(3²¹⁶+3²¹⁷+3²¹⁸+3²¹⁹+3³²⁰+3³²¹+3³²²+3³²³)

S=(3+3²+3³+3⁴+3⁵+3⁶+3⁷+3⁸)+....+3²¹⁵.(3+3²+3³+3⁴+3⁵+3⁶+3⁷+3⁸)

S=9840+...+3²¹⁵.9840

S=9840.(1+...+3²¹⁵)

S=41.240.(1+...+3²¹⁵)\(⋮\)41

Vậy S\(⋮\)41

Chúc bn học tốt

Nguyễn Trí Nghĩa (Team ngọc rồng) đề bài không có sai đâu bạn đề bài đúng đấy cô giáo mk cx cho bài này mak

Ta có :

 \(S=4+3^2+3^3+.....+3^{223}\)

\(=1+3+3^2+3^3+....+3^{223}\)

\(\Rightarrow3S=3+3^2+3^3+3^{224}\)

\(\Leftrightarrow S=\frac{3^{224}-1}{2}=\frac{\left(3\right)^{4^{56}}-1}{2}\)

Vì  \(3^4\equiv-1\left(mod41\right)\)

\(\Rightarrow3^{4^{56}}\equiv1\left(mod41\right)\)

\(\Leftrightarrow3^{4^{56}}-1\equiv0\left(mod41\right)\)

\(\Leftrightarrow\frac{3^{4^{56}}-1}{2}\equiv0\left(mod41\right)\)

Hay \(S⋮41\) ( đpcm )

B = (1 + 3) + (3²+3³)+.....+(3⁸⁹+3⁹⁰)

  = 4 + 3² .(1 + 3) + .....+3⁹⁰.(1+3)

 = 1 .4 + 3².4 + ..... +3⁹⁰.4

= 4.(1 + 3² + .... +3⁹⁰) chia hết cho 4

\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)

Lời giải:
$S=(2+2^2)+(2^3+2^4)+....+(2^{23}+2^{24})$

$=2(1+2)+2^3(1+2)+....+2^{23}(1+2)$

$=(1+2)(2+2^3+...+2^{23})$

$=3(2+2^3+...+2^{23})\vdots 3$

b.

$S=2+2^2+2^3+...+2^{23}+2^{24}$

$2S=2^2+2^3+2^4+....+2^{24}+2^{25}$

$\Rightarrow 2S-S=2^{25}-2$

$\Rightarrow S=2^{25}-2$

Ta có:

$2^{10}=1024=10k+4$

$\Rightarrow 2^{25}-2=2^5.2^{20}-2=32(10k+4)^2-2=32(100k^2+80k+16)-2$
$=10(320k^2+8k+51)\vdots 10$

$\Rightarrow S$ tận cùng là $0$

\(S=\left(3+3^{3+3^3}\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\)

\(S=39.1+39.3^3+....+39.3^{96}=>S=39\left(1+3^3+3^6+.....+3^{96}\right)\)

Vậy S chia hết cho 39