cho C = 5+5^2+5^3+5^4+..+5^2018 . chung to:C chia het cho 126
cho S= 5+5^2+5^3...+5^2006
a, tinh S
b,Chung minh rang S chia het cho 126
cho S=5+52+53+..........+52006
a,Tinh S
b,Chung minh S chia het cho 126
S = 5 + 52 + 53 + ......... + 52006
5S = 52 + 53 + 54 + .......... + 52007
5S - S = ( 52 + 53 + 54 + .......... + 52007) - ( 5 + 52 + 53 + ......... + 52006 )
4S = 52007 - 5
S = \(\frac{5^{2007}-5}{4}\)
a)\(S=5+5^2+5^3+.....+5^{2006}\Rightarrow5S=5^2+5^3+5^4+\)\(....+5^{2007}\)
\(\Rightarrow5S-S=\left(5^2+5^3+5^4+....+5^{2007}\right)-\left(5+5^2+5^3+.....+5^{2006}\right)\)
\(\Rightarrow4S=5^{2007}-5\Rightarrow S=\frac{5^{2007}-5}{4}\)
\(a.S=5+5^2+5^3+......+5^{2006}\)
\(S=\left(5+5^2+5^3+5^4+5^5+5^6\right)+.....+\left(5^{2001}+5^{2002}+.....+5^{2006}\right)\)
\(S=5.\left(1+5+5^2+5^3+5^4+5^5\right)+......+5^{2001}\left(1+5+5^2+5^3+5^4+5^5\right)\)
\(S=5.3906+........+5^{2001}.3906\)
\(S=3906\left(5+....+5^{2001}\right)\)
\(b.S=3906\left(5+....+5^{2001}\right)\)
\(S=126.3\left(5+....+5^{2001}\right)\)
\(\Rightarrow\text{S chia hết cho 126}\)
bai 1; cho tong M =126 +213+x. Tim x de M chia het cho 3
bai 2; chung to rang tong ; A= 2 + 2 mu 3 + 2 mu 4 + 2 mu 5 + 2 mu 6 +2 mu 7 +2 mu 9 + 2 mu 10 + 2 mu 12 chia het cho 5
Có : 126 chia hết cho 3, 213 chia hết cho 3
Để được M chia hết cho 3 thì x phải chia hết cho 3
Hay gọi là 3k ( k thuộc N)
2.
Hình như đầu bài bài 2 sai
dung do khong sai dau
chứng minh
a ) 5^5 - 5^4 + 5^3 chia het cho 7
b) 3 ^n+2 - 2^n+2 + 3^n - 2^n chia het cho 10
c) 3 ^n+3 + 3^n+1 + 2^+3 + 2^n+2 chia het cho 6
d ) A = 2+2^2+2^3+....+ 2^12 chia het cho 7
g ) B= 2^35 + 2^36 + 2^37 + 2^38 chia het cho 3
k) C = 1 + 3 + 3^2 + ...+ 3^61
chung to C chia het cho 4
chung to C k chia het cho 3
h ) 5^n+2 + 3^n+2 - 3^n - 5^n chia het cho 24
gíúp mk vs ạ
Cho S = 5+52 + 53 +......+596
a, chung minh rang : S chia het cho 126
S = 5^2 + 5^3 + 5^4 + ... + 5^2006 chia het cho 126
S = 5 + 52 + 53 +....+52006
S= (5+52+53+54+55+56) +.....+ ( 22001+52002+52003+52004+52005+52006)
S= 5 x ( 1+5+52+53+5455 ) +......+ 52001x (1+5+5 2+53+54+55)
S= 5 x 3906+.........+ 52001 x 3906
S = 3906x( 5+..+52001)
S = 3906 x ( 5+...+52001)
S = 126 x 3 x ( 5+...+52001)
=> S chia hết 126
Cho S = 5+5^2+5^3+5^4+5^5+5^6+..+5^2004. Chứng minh S chia het cho 126 va 65
1.Chung minh tong 2+22+23+24+......+220 chia het cho 5
2.Tim so tu nhien n de 2n+5 chia het cho n+1
3. Cho S=5+52+53+54+55+56+......+52012
chung minh S chia het cho 65
minh dang can gap
CMR : 5+5^2+5^3+... +5^96 chia het cho 126
có:2B+3=3+2B=3+2*3+2*3²+2*3^3+2*3^4+.......
=3²+2*3²+2*3^3+2*3^4+....+2*3^100=3^3+...
=......=3^101
3^x=3^101 <=> x=101
5 + 52 + 53 + ... + 596 (có 96 số, 96 chia hết cho 6)
= (5 + 52 + 53 + 54 + 55 + 56) + (57 + 58 + 59 + 510 + 511 + 512) + ... + (591 + 592 + 593 + 594 + 595 + 596)
= (5 + 54) + (52 + 55) + (53+ 56) + ... + (591 + 594) + (592 + 595) + (593+ 596)
= 5.(1 + 53) + 52.(1 + 53) + 53.(1 + 53) + ... + 591.(1 + 53) + 592.(1 + 53) + 593.(1 + 53)
= 5.126 + 52.126 + 53.126 + ... + 591.126 + 592.126 + 593.126
= 126.(5 + 52 + 53 + ... + 591 + 592 + 593) chia hết cho 126
=> đpcm