B=(3+3^5)+(3^2+3^6)+...+(3^1987+3^1991)

B=3*(1+3^4)+3^2*(1+3^4)+...+3^1987*(1+3^4)

B=3*82+3^2*82+...+3^1987*82

B=82*(3+3^2+...+3^1987)

B=41*2*(3+3^2+...+3^1987)

Nên B chia hết cho 41

E = 1 + 3 + 3² + 3³ +.....+3¹¹⁹

E = ( 1 + 3 + 3²) +....+ ( 3¹¹⁷ + 3¹¹⁸+ 3¹¹⁹)

E =   13 + ......+ 3¹¹⁷.( 1 + 3 + 3²)

E = 13 +.....+ 3¹¹⁷ . 13

E = 13. ( 3⁰ + ....+ 3¹¹⁷)

13 ⋮ 13 ⇒ 13. (3⁰ +....+3¹¹⁷) ⋮ 13 ⇒ E = 1 +3+3²+ ....+3¹¹⁹⋮13(đpcm)

=\(\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)

= \(13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

=\(13+3^3.13+...+3^{117}.13\)

=\(13.\left(1+3^2+...+3^{117}\right)\) chia hết cho 13

Do A có 30 số hạng, ta nhóm 3 số thành 1 nhóm nên vừa đủ 10 nhóm và không dư số nào.

A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^30

= (2+2^2+2^3)+(2^4+2^5+2^6)+...+(2^28+2^29+2^30)

= 2(1+2+2^2)+2^4(1+2+2^2)+...+2^28(1+2+2^2)

= 2.7 + 2^4 .7 + ... + 2^28 .7

= 7(2+2^4+...+2^28) chia hết cho7 (DPCM)

A = 2 + 2^2 + 2^3 + 2^4 + ... + 2^30

= (2+2^2+2^3)+...+(2^28+2^29+2^30)

= 2(1+2+2^2)+...+2^28(1+2+2^2)

= 2.7 + ... + 2^28 .7

= 7.(2+...+2^28) chia hết cho 7

a) \(B=3+3^2+...+3^{90}\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+\left(3^{89}+3^{90}\right)\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+3^{88}.\left(3+3^2\right)\)

\(\Leftrightarrow B=12+...+3^{88}.12\)

\(\Leftrightarrow B=12.\left(1+...+3^{88}\right)⋮4\left(đpcm\right)\)

b)\(B=3+3^2+...+3^{90}\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+\left(3^{89}+3^{90}\right)\)

\(\Leftrightarrow B=\left(3+3^2\right)+...+3^{88}.\left(3+3^2\right)\)

\(\Leftrightarrow B=12+...+3^{88}.12\)

\(\Leftrightarrow B=12.\left(1+...+3^{88}\right)⋮12\left(đpcm\right)\)

c) \(B=3+3^2+...+3^{90}\)

\(\Leftrightarrow B=\left(3+3^2+3^3\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(\Leftrightarrow B=\left(3+3^2+3^3\right)+...+3^{87}.\left(3+3^2+3^3\right)\)

\(\Leftrightarrow B=39+...+3^{87}.39\)

\(\Leftrightarrow B=39.\left(1+..+3^{87}\right)⋮39\left(đpcm\right)\)

\(B=2^2+2^3+2^4+...+2^{121}\\=(2^2+2^3)+(2^4+2^5)+(2^6+2^7)+...+(2^{120}+2^{121})\\=2^2\cdot(1+2)+2^4\cdot(1+2)+2^6\cdot(1+2)+...+2^{120}\cdot(1+2)\\=2^2\cdot3+2^4\cdot3+2^6\cdot3+...+2^{120}\cdot3\\=3\cdot(2^2+2^4+2^6+...+2^{120})\)

Vì \(3\cdot(2^2+2^4+2^6+...+2^{120})\vdots3\)

nên \(B\vdots3\)

Bài 1 :

Ta có :

a chia 3 dư 1 \(\Rightarrow a=3k+1\)

b chia 3 dư 2 \(\Rightarrow b=3k_1+2\) \(\left(k;k_1\in N\right)\)

\(ab=\left(3k+1\right)\left(3k_1+2\right)=3k.k_1+2.3k+3.k_1+2\)

Mà \(3k.k_1+2.3k+3.k_1⋮3\)

\(\Rightarrow3k.k_1+2.3k+3.k_1+2\) chia 3 dư 2

\(\Rightarrow ab\) chia 3 dư 2 \(\rightarrowđpcm\)

Bài 2 :

Ta có :

\(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
\(=-5n⋮5\)

\(\Rightarrow n\left(2n-3\right)-3n\left(n+1\right)⋮5\) với mọi n

\(\rightarrowđpcm\)

a,n²+3n+3 chia hết cho n+1

=>n²+n+2n+2+1 chia hết cho n+1

=>n(n+1)+2(n+1)+1 chia hết cho n+1

=>1 chia hết cho n+1

=>n+1 E Ư(1)={1;-1}

=>n E {0;-2}

b, n²+4n+2 chia hết cho n+2

=>n²+2n+2n+4-2 chia hết cho n+2

=>n(n+2)+2(n+2)-2 chia hết cho n+2

=>2 chia hết cho n+2

=>n+2 E Ư(2)={1;-1;2;-2}

=>n E {-1;-3;0;-4}

c, n²-2n+3 chia hết cho n-1

=>n²-n-n+1+4 chia hết cho n-1

=>n(n-1)-(n-1)+4 chia hết cho n-1

=>4 chia hết cho n-1

=>n-1 E Ư(4)={1;-1;2;-2;4;-4}

=>n E {2;0;3;-1;5;-3}

Cảm ơn nha ko có bạn chắc thầy cắt tiết mik rùi

Thế S là số nào bn mà chia hết cho 3 vậy bn ?