cho s=1+3+3^2+3^3+…+3^39
a)tính s
b)chứng minh rằng s chia het cho 4 va40
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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
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
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 \(2^3\equiv-1\left(mod9\right)=>2^{204}\equiv1\left(mod3\right)\)
=>\(=>2^{204}-1⋮9=>3S⋮9=>S⋮3\left(ĐPCM\right)\)
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
a, S = 1+3+32+33+...+339
3S = 3+32+33+34+...+340
2S = 3S - S = 340 - 1
=> S = \(\frac{3^{40}-1}{2}\)
b, S = 1+3+32+33+...+339
S = (1+3+32+33)+(34+35+36+37)+....+(336+337+338+339)
S = 1(1+3+32+33) + 34(1+3+32+33) +.....+ 336.(1+3+32+33)
S = 1.40 + 34.40 +.....+ 336.40
S = 40.(1+34+...+336) chia hết cho 40
S = 4.10.(1+34+...+336) chia hết cho 4
=> Đpcm