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2S=2^2+2^3+2^4+...+2^61
2S-S=S=2^61-2
còn câu b bạn tự làm nhé
a/ Ta có :
\(S=1+3+3^2+........+3^{2017}\)
\(\Leftrightarrow S=\left(1+3\right)+\left(3^2+3^3\right)+......+\left(3^{2016}+3^{2017}\right)\)
\(\Leftrightarrow S=1\left(1+3\right)+3^2\left(1+3\right)+......+3^{2016}\left(1+3\right)\)
\(\Leftrightarrow S=1.4+3^2.4+........+3^{2016}.4\)
\(\Leftrightarrow S=4\left(1+3^2+......+3^{2016}\right)⋮4\left(đpcm\right)\)
b/ \(S=1+3+..........+3^{2017}\)
\(\Leftrightarrow3S=3+3^2+.........+3^{2017}+3^{2018}\)
\(\Leftrightarrow3S-S=\left(3+3^2+..........+3^{2018}\right)-\left(1+3+.....+3^{2017}\right)\)
\(\Leftrightarrow2S=3^{2018}-1\)
\(\Leftrightarrow S=\dfrac{3^{2018}-1}{2}\)
a, \(S=3^0+3^2+3^4+...+3^{2002}\)
\(3^2S=3^2+3^4+3^6+...+3^{2004}\)
\(8S=3^{2004}-1\)
\(S=\frac{3^{2004}-1}{8}\)
S=30+32+34+36+...+32002
S=(30+32+34)+(36+38+310)+...(31988+32000+32002)
S=91.1+36.(30+32+34)+...+31988.(30+32+34)
S=91.1+36.91+...+31988.91
S=91.(1+36+..+31988)
S=7.13.(1+36+..+31988)
=>S chia hết cho 7
a) \(S=1+3^2+3^4+3^6+...+3^{2002}\)
\(3^2.S=3^2+3^4+3^6+3^8+...+3^{2004}\)
\(9S-S=\left(3^2+3^4+3^6+3^8+...+3^{2004}\right)-\left(1+3^2+3^4+3^6+...+3^{2002}\right)\)
\(8S=3^{2004}-1\)
\(S=\frac{3^{2004}-1}{8}\)
b) \(S=1+3^2+3^4+3^6+...+3^{2002}\)
\(=\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+...+2^{1998}\left(1+3^2+3^4\right)\)
\(=\left(1+3^2+3^4\right)\left(1+3^6+...+3^{1998}\right)\)
\(=91\left(1+3^6+...+3^{1998}\right)\)
\(=7.13\left(1+3^6+...+3^{1998}\right)\)
Vậy S chia hết cho 7
a,S=\(\left(2+2^2+2^3+2^4+2^5+2^6\right)+.....+\left(2^{85}+2^{86}+2^{87}+2^{88}+2^{89}+2^{90}\right)\)
\(=126+2^6.\left(2+2^2+2^3+2^4+2^5+2^6\right)+...+2^{84}.\left(2+2^2+2^3+2^4+2^5+2^6\right)\)
\(=126+2^6.126+...+2^{84}.126\)
\(=126.\left(2^0+2^6+2^{12}+....+2^{84}\right)=21.6.\left(2^0+2^6+....+2^{84}\right)\) chia hết cho 21
b,Xét x=0 thì \(5^y=1+124=125\Rightarrow y=3\)(thỏa mãn)
Xét x\(>0\) thì \(5^y>1+124=125>0\) nên \(5^y\) là số lẻ mà \(2^x\) là số chẵn \(\Rightarrow2^x+124\) là số chẵn(vô lí)
Vậy x=0,y=3 thỏa mãn
S = 1 - 3 + 3^2 - 3^3 + ... + 3^98 - 3^99
S = (1 - 3 + 3^2 - 3^3) + ... + (3^96 - 3^97 + 3^98 - 3^99 )
S = (-20) + (-20) +...+ (-20) (24 số -20)
S = (-20).24 chia hết cho -20
=> đpcm
Câu hỏi của Nguyễn Dương - Toán lớp 6 - Học toán với OnlineMath
Bạn tham khảo.
a, S=1+2^7+(2+2^2)+(2^3+2^4)+(2^5+2^6)
S=1+128+2*3+(2^3*1+2^3*2)+(2^5*1+2^5*2)
S=129+2*3+2^3*(1+2)+2^5*(1+2)
S=3*43+2*3+2^3*3+2^5*3
S=3*(43+2+2^3+2^5)chia hết cho 3 nên S chia hết cho 3
c) S = ( -2 ) + 4+ ( -6 ) + 8 + ... + ( -2002 ) + 2004
S = [ (-2)+4] + [ (-6) + 8 ] + ... + [ (-2002) + 2004 ]
S = 2 + 2 + 2 + ... + 2 ( 501 số hạng 2 )
S = 2*501
S = 1002
Lời giải:
$S=3^0+3^2+3^4+...+3^{2014}$
$3^2S=3^2+3^4+3^6+...+3^{2016}$
$\Rightarrow 3^2S-S=3^{2016}-3^0$
$\Rightarrow 8S=3^{2016}-1$
$\Rightarrow S=\frac{3^{2016}-1}{8}$
b.
$S=(3^0+3^2+3^4)+(3^6+3^8+3^{10})+....+(3^{2010}+3^{2012}+3^{2014})$
$=(1+3^2+3^4)+3^6(1+3^2+3^4)+...+3^{2010}(1+3^2+3^4)$
$=(1+3^2+3^4)(1+3^6+...+3^{2010})=91(1+3^6+...+3^{2010})$
$=7.13(1+3^6+...+3^{2010})\vdots 7$.
a ) S = 20 +22 + 24 +...+ 22014
4S = 22 + 24 + 26 + ... + 22016
Mà S = ( 4S- S ) : 3
=> S = [ ( 22 + 24 + 26 +...+ 22016 ) - ( 20 + 22 + 24 +...+ 22014 ) ] : 3
= [ 22016 - 20 ] : 3
= \(\frac{2^{2016}-1}{3}\)
b) S = 20 + 22 + 24 + ... + 22014
= ( 20 + 22 + 24 ) + ( 25 + 26 + 27 ) + ...+ ( 22010 + 22012 + 22014 )
= 21 + 25 x ( 20 + 22 + 24 ) +... + 22010 x ( 20 + 22 + 24 )
= 21 + 25 x 21 + ... + 22010 x 21
= 21 x ( 1 + 25 + ... + 22010 )
=> S \(⋮\)21 (đpcm)