Lời giải:

$S=3^2+3^4+3^6+...+3^{998}+3^{1000}$

$3^2S=3^4+3^6+3^8+...+3^{1000}+3^{1002}$

$\Rightarrow 3^2S-S=3^{1002}-3^2$
$\Rightarrow 8S=3^{1002}-9$

$\Rightarrow S=\frac{3^{1002}-9}{8}$

b.

$S=3^2+3^4+(3^6+3^8+3^{10})+(3^{12}+3^{14}+3^{16})+...+(3^{996}+3^{998}+3^{1000})$

$=90+3^6(1+3^2+3^4)+3^{12}(1+3^2+3^4)+...+3^{996}(1+3^2+3^4)$

$=90+(1+3^2+3^4)(3^6+3^{12}+...+3^{996})$

$=90+91(3^6+3^{12}+...+3^{996})$

$=6+ 12.7+7.13(3^6+3^{12}+...+3^{996})$ chia $7$ dư $6$

B = (1 + 3) + (3²+3³)+.....+(3⁸⁹+3⁹⁰)

  = 4 + 3² .(1 + 3) + .....+3⁹⁰.(1+3)

 = 1 .4 + 3².4 + ..... +3⁹⁰.4

= 4.(1 + 3² + .... +3⁹⁰) chia hết cho 4

\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)

S = 3¹⁰⁰ - 1

Xét \(2^2S=2^2+2^4+.....+2^{204}\)

=>\(\left(2^2-1\right)S=2^{204}-2^0\)

=>3S=\(2^{204}-1\)

Ta có \(2^3\equiv-1\left(mod9\right)=>2^{204}\equiv1\left(mod3\right)\)

=>\(=>2^{204}-1⋮9=>3S⋮9=>S⋮3\left(ĐPCM\right)\)

thank

Bạn vào câu hỏi tương tự nha !!!

ta có:\(S_3=1+3+3^2+3^3+...+3^{11}=\left(1+3+3^2+3^3\right)+...+3^8+3^9+3^{10}+3^{11}\)

\(=1+3+3^2+3^3+..+3^8.\left(1+3+3^2+3^3\right)=40+..+3^8.40=40.\left(..\right)\) chia hết cho 40