Ta có: B= 30 + 32 + ... + 32002.
B = \(\left(1+3^2+3^4\right)+...+\)\(\left(3^{1998}+3^{2000}+3^{2002}\right)\)
B = \(\left(1+9+81\right)+...+3^{1998}\left(1+3^2+3^4\right)\)
B = \(91+...+3^{1998}.91\)
B = \(91\left(1+3^6+...+3^{1998}\right)\)
B= \(7.13\left(1+3^6+...+3^{1998}\right)\)
Do 7.13 ( 1 + 3^6 + ... + 3^1998) \(⋮\)7
=> B chia hết cho 7
\(B=3^0+3^2+...+3^{2002}\)
\(B=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{1998}+3^{2000}+3^{2002}\right)\)
\(B=91+3^6\left(3^0+3^2+3^4\right)+...+3^{1998}\left(3^0+3^2+3^4\right)\)
\(B=91+3^6.91+...+3^{1998}.91\)
\(B=91\left(1+3^6+...+3^{1998}\right)\)
VÌ \(91⋮7\Rightarrow91\left(1+3^6+...+3^{1998}\right)⋮7\)
NÊN \(B⋮7\)
VẬY, \(B\)\(⋮\)\(7\)
NHÉ K CHO MÌNH NHÉ !
B=(30+32+34)+...+(31998+32000+32002)
B=91+...+31998. · 91
B=13·7+...+31998 · 13 · 7 \(⋮\)7
Vậy B chia hết cho 7
