a) \(S=3^0+3^2+3^4+3^6+...+3^{2002}\)
\(\Rightarrow S=1+3^2+3^4+...+3^{2002}\)
\(\Rightarrow9S=3^2+3^4+3^6+...+3^{2004}\)
\(\Rightarrow9S-S=\left(3^2+3^4+3^6+...+3^{2004}\right)-\left(1+3^2+3^4+...+3^{2002}\right)\)
\(\Rightarrow8S=3^{2004}-1\)
\(\Rightarrow S=\frac{3^{2004}-1}{8}\)
b) \(S=3^0+3^2+3^4+3^6+...+3^{2002}\)
\(\Rightarrow S=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2000}+3^{2001}+3^{2002}\right)\)
\(\Rightarrow S=\left(1+9+81\right)+3^6.\left(1+3^2+3^4\right)+...+3^{2000}.\left(1+3^2+3^4\right)\)
\(\Rightarrow S=91+3^6.91+...+3^{2000}.91\)
\(\Rightarrow S=\left(1+3^6+...+3^{2000}\right).91⋮7\)
\(\Rightarrow S⋮7\)
b) Câu này mình có cách khác:
Ta có S là số nguyên nên phải chứng minh \(3^{2004}-1\) chia hết cho 7
Ta có: \(3^{2004}-1=\left(3^6\right)^{334}-1=\left(3^6-1\right).M=728.M=7.104.M\)
\(\Rightarrow3^{2004}\) chia hết cho 7. Mặt khác \(\left(7;8\right)=1\) nên S chia hết cho 7