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Nhân S với 3^2 ta được 9S=3^2+3^4+....+3^2002+3^2004
=>9S-S=(3^2+3^4+....+3^2004)-(3^0+3^2+....+3^2002)
=>8S=3^2004-1
=>S=(3^2004-1)/8
b,ta có S là sô nguyên nên fải chung minh 3^2004-1chia hết cho 7
ta có : 3^2004-1=(3^6)^334-1=(3^6-1).M=728.M=7.104.M
=>3^2004 chia hết cho 7. Mặt khác (7;8)=1 nên S chia hết cho 7
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 $6$
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
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
Tham khảo :
a, S=30+32+34+36+...+32020
⇔32S=32+34+36+38+...+32022
⇔32S−S=32022−30
⇔9S−S=32022−1
⇔8S=32022−1⇔S=32022−18
b,S=30+32+34+36+...+32020
=(30+32+34)+(36+38+310)+...+(32016+32018+32020)
=(1+32+34)+36(1+32+34)+...+32016(1+32+34)
=(1+32+34)(1+36+...+32016)
=91(1+36+...+32016)=13.7(1+36+...+32016)⋮7 (
=> (đpcm)
=>99
s = 3 ^0 + 3 ^ 2 + 3^ 4+ 3 ^6 +... + 3 ^2002
9S = 3 ^4 + 3^6 + 3 ^ 2004
9S - S= 3 ^ 2004 - 1
8S = 3^2004 - 1
S = 3 ^ 2004 - 1/8
k mk nha
a)nhân S với 32 ta dc:
9S=3^2+3^4+...+3^2002+3^2004
=>9S-S=(3^2+3^4+...+3^2004)-(3^0+3^4+...+2^2002)
=>8S=32004-1
=>S=32004-1/8
b) ta có S là số nguyên nên phải chứng minh 32004-1 chia hết cho 7
ta có:32004-1=(36)334-1=(36-1).M=7.104.M
=>32004 chia hết cho 7. Mặt khác ƯCLN(7;8)=1 nên S chia hết cho 7
cho tổng :S=3^0+3^2+3^4+3^6+...........................+3^2014.tính S và chứng minh S chia hết cho 7
\(S=3^0+3^2+3^4+3^6+...+3^{2014}\)
\(=1+3^2+3^4+3^6+...+3^{2014}\)
\(=\left(1+3^2\right)+3^4\left(1+3^2\right)+...+3^{2012}\left(1+3^2\right)\)
\(=7+3^4.7+...+3^{2012}.7=7\left(1+3^4+...+3^{2012}\right)⋮7\)
Vậy ta có đpcm