Question

cho : S = 3⁰ + 3² + 3⁴ + 3⁶ + ...........+ 3²⁰⁰²

^{a ) thu gọn S} 

^{b ) chứng tỏ S chia hết cho 7}

Answer 1

a)  \(S=1+3^2+3^4+3^6+...+3^{2002}\)

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

\(9S-S=\left(3^2+3^4+3^6+3^8+...+3^{2004}\right)-\left(1+3^2+3^4+3^6+...+3^{2002}\right)\)

\(8S=3^{2004}-1\)

\(S=\frac{3^{2004}-1}{8}\)

 

b)  \(S=1+3^2+3^4+3^6+...+3^{2002}\)

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

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

\(=91\left(1+3^6+...+3^{1998}\right)\)

\(=7.13\left(1+3^6+...+3^{1998}\right)\)

Vậy S chia hết cho 7