Question

Cho S=3 mũ 0+3 mũ 2+3 mũ 4 +3 mũ 6 +.....+3 mũ 2020

a)Tính S

b)Chứng minh S chia hết cho 7

Answer 1

a, \(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(\Leftrightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{2022}\)

\(\Leftrightarrow3^2S-S=3^{2022}-3^0\)

\(\Leftrightarrow9S-S=3^{2022}-1\)

\(\Leftrightarrow8S=3^{2022}-1\Leftrightarrow S=\frac{3^{2022}-1}{8}\)

b,\(S=3^0+3^2+3^4+3^6+...+3^{2020}\)

\(=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2016}+3^{2018}+3^{2020}\right)\)

\(=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{2016}\left(1+3^2+3^4\right)\)

\(=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{2016}\right)\)

\(=91\left(1+3^6+...+3^{2016}\right)=13.7\left(1+3^6+...+3^{2016}\right)⋮7\)

=> đpcm

Answer 2

Tham khảo :

a, $S=3^0+3^2+3^4+3^6+...+3^{2020}$

$\iff 3^{2}S=3^2+3^4+3^6+3^8+...+3^{2022}$

$\iff 3^{2}S-S=3^{2022}-3^0$

$\iff 9S-S=3^{2022}-1$

$\iff 8S=3^{2022}-1\iff S=\frac{3^{2022}-1}{8}$

b,$S=3^0+3^2+3^4+3^6+...+3^{2020}$

$=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{2016}+3^{2018}+3^{2020}\right)$

$=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+3^{2016}\left(1+3^2+3^4\right)$

$=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{2016}\right)$

$=91\left(1+3^6+...+3^{2016}\right)=13.7\left(1+3^6+...+3^{2016}\right)⋮7$

=> (đpcm)