Question

Cho : \(S=3^0+3^2+3^4+3^6+...+3^{2002}\)

a. Tính S

b. Chứng minh rằng S : 7

 

Answer 1

xin lỗi chị em mới học lớp 5

Answer 2

b; S=(3^0+3^2+3^4)+......+(3^1998+3^200+3^202)

       =91+.....3^1998*(1+3^2+3^4)

       =91+.....+3^1998*91

       =91+.....+3^1998*13*7 => S chia het cho 7

Answer 3

a) \(9S=3^2+3^4+3^6+3^8+...+3^{2004}\)

\(9S-S=\left(3^2+3^4+3^6+3^8+...+3^{2004}\right)-\left(3^0+3^2+3^4+3^6+...+3^{2002}\right)\)

\(9S-S=3^2+3^4+3^6+3^8+...+3^{2004}-3^0-3^2-3^4-3^6-...-3^{2002}\)

\(8S=3^{2004}-3^0\)

\(S=3^{2004}-3^0:8\)

b) \(S=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{1998}+3^{2000}+3^{2002}\right)\)

\(S=3^0.\left(1+3^2+3^4\right)+3^6.\left(1+3^2+3^4\right)+...+3^{1998}.\left(1+3^2+3^6\right)\)

\(S=3^0.91+3^6.91+...+3^{1998}.91\)

\(S=91.\left(3^0+3^6+...+3^{1998}\right)\)

\(S=\left(13.7\right).\left(3^0+3^6+...+3^{1998}\right)\)

\(\Rightarrow\) S chia hết cho 7.