a) B = 3 + 32 + 33 + ... + 360
=(3+32)+(33+34)+...+(359+360)
=3(1+3)+33(1+3)+...+359(1+3)
=(3+1)(3+33+...+359)
=4(3+33+...+359)
=>B chia hết cho 4
câu a trước nè **** caj làm típ
b) B=(3+32+33)+...+(358+359+360)
=30(3+32+33)+...+357(358+359+360)
=3+32+33(30+33+36+...+357)
=39(30+33+36+...+357) chia hết cho 13
Vậy B chia hết cho 13
**** cả 2 bài nha
a) B = 3 + 32 + 33 + ... + 360
=(3+32)+(33+34)+...+(359+360)
=3(1+3)+33(1+3)+...+359(1+3)
=(3+1)(3+33+...+359)
=4(3+33+...+359)
=>B chia hết cho 4
b)B = 3 + 32 + 33 + ... + 360
=(3+32+33)+(34+35+36)+...+(358+359+360)
=3(1+3+32)+34(1+3+32)+...+358(1+3+32)
=3.(1+3+9)+34(1+3+9)+...+358(1+3+9)
=(1+3+9)(3+34+...+358)
=13(3+34+...+358)
=>B chia hết cho 13
A=3+3^2+3^3+...+3^25
Chứng tỏ rằng A không chia hết cho 39
Cho A= 3+3^2+3^3+ 3^4+...+3^60
Hãy chứng minh A chia cho 13
a,\(B=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{59}.4\)
\(=4\left(3+3^3+...+3^{59}\right)⋮4\)
\(\Rightarrow B⋮4\)
b,\(B=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=3.13+...+3^{58}.13\)
\(=13\left(3+3^4+...+3^{58}\right)⋮13\)
\(\Rightarrow B⋮13\)
