Question

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

a.tính S

b.CMR:S chia hết cho 7

Answer 1

Easy????

a) Ta có: S = \(3^0+3^{2^{ }}+...+3^{2002}\)

=> 3²S = \(3^2+3^4+3^6+...+3^{2004}\)

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

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

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

b) Ta lại có: S = \(3^0+3^{2^{ }}+...+3^{2002}\)

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

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

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

Vì 91 \(⋮\) 7 => \(91\left(3^0+3^6+...+3^{1998}\right)\) \(⋮\) 7

=> S \(⋮\) 7 ( đpcm)