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A=(3+3^2+3^3+3^4)+.....+(3^97+3^98+3^99+3^100)
=3(1+3+3^2+3^3)+3^2(1+3+3^2+3^3)+....+3^97(1+3+3^2+3^3)
=3.40+3^2.40++....+3^97.40 chia hêt cho 40
câu này dễ mà
A= 3+3^2+3^3+3^4+..........+3^100.
A= (3+3^2+3^3+3^4)+...+(3^97+3^98+3^99+3^100) ( nhóm 4 số lại)
A= 3(1+3+3^2+3^3)+...+ 3^97(1+3+3^2+3^3) ( rút ra)
A=3x40+3^5x40+...+3^97x40
A=40(3+3^5+..+3^97) chia hết cho 40
ĐẶt A=3+3^2+3^3+....+3^100
A= 3(1+3+3^2+3^3) +3^5(1+3+3^2+3^3)+...... + 3^97 ( 1 + 3 + 3^2 + 3^3)
A=3.40 +3^5.40+.....+3^97.40
Vì 40 chia hết cho 40 => 3.40 +3^5.40+.....+3^97.40
Vậy A chia hết cho 40.
A=3+32+33+34+...+3100 chia hết cho 40
A=(3+32+33+34)+(35+36+37+38)+...+(397+398+399+3100)
A=3.(1+3+32+33)+35.(1+3+32+33)+...+397.(1+3+32+33)
A=3.40+35.40+...+397.40
A=40.(3+35+...+397) chia hết cho 40 (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
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
ĐẶt A=3+3^2+3^3+....+3^100
Chia A thành từng nhóm 4 số (vì A có 100 số) ta được 25 nhóm
A= 3(1+3+3^2+3^3) +3^5(1+3+3^2+3^3)+......
+3^97(1+3+3^2+3^3)
A=3.40 +3^5.40+.....+3^97.40
Vậy A chia hết cho 40.
C=\(3+3^2+3^3+3^4+...+3^{100}\)
= \(3\times\left(1+3+3^2+3^3+...+3^{99}\right)\)
= \(3\times\left[\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\right]\)
= \(3\times\left(40+3^4\times40+...+3^{96}\times40\right)\)
= \(3\times40\times\left(1+3^4+...+3^{96}\right)\)chia het cho 40
=> C chia het cho 40
\(3+3^2+3^3+3^4+...+3^{100}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=1\left(3+3^2+3^3+3^4\right)+...+3^{96}\left(3+3^2+3^3+3^4\right)\)
\(=\left(1+...+3^{96}\right)\left(3+3^2+3^3+3^4\right)\)
\(=Q.120\)
\(=Q.3.40\)
\(\Rightarrow3+3^2+3^3+3^4+...+3^{100}⋮40\) (Đpcm)
A = 1 + 3^1 + 3^2 + ... + 3^99
3A = 3 + 3^2+ 3^3 + ... + 3^100
3A = ( 3 + 3^2 ) + ( 3^3 + 3^4 ) + ... + ( 3^99 + 3^100 )
3A = 3 ( 1 + 3 ) + 3^3 ( 1 + 3 ) + ... + 3^99 ( 1 + 3 )
3A = 3 . 4 + 3^3 . 4 + ... + 3^99 . 4
3A = 4 . ( 3 + 3^3 + 3^99 ) \(⋮\)4
help mình!!!!!plz
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1
Gọi d = ƯCLN(2n + 5; 3n + 7) (với d ∈N*)
\(\Rightarrow\hept{\begin{cases}2n+5⋮d\\3n+7⋮d\end{cases}}\) \(\Rightarrow\hept{\begin{cases}3\left(2n+5\right)⋮d\\2\left(3n+7\right)⋮d\end{cases}}\) \(\Rightarrow\hept{\begin{cases}6n+15⋮d\\6n+14⋮d\end{cases}}\)
\(\text{⇒ (6n + 15) – (6n + 14) ⋮ d}\)
\(\text{⇒1 ⋮d}\)
\(\text{⇒d = 1}\)
Do đó: \(\text{ƯCLN(2n + 5; 3n + 7) = 1}\)
Vậy hai số \(\text{2n + 5 và 3n +7 }\)là hai số nguyên tố cùng nhau.
\(M=1+3+3^2+...+3^{100}\)
\(\Leftrightarrow M=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(\Leftrightarrow M=4+3^2+\left(1+3+3^2\right)+3^5+\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(\Leftrightarrow M=4+3^2.13+3^5.13+...+3^{98}.13\)
\(\Leftrightarrow M=4+13\left(3^2+3^5+...+3^{98}\right)\)
mà \(13\left(3^2+3^5+...+3^{98}\right)⋮13\)
\(4:13\left(dư4\right)\)
\(\Leftrightarrow M:13\left(dư4\right)\)
