![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)\(S=5+5^2+5^3+.....+5^{2006}\Rightarrow5S=5^2+5^3+5^4+\)\(....+5^{2007}\)
\(\Rightarrow5S-S=\left(5^2+5^3+5^4+....+5^{2007}\right)-\left(5+5^2+5^3+.....+5^{2006}\right)\)
\(\Rightarrow4S=5^{2007}-5\Rightarrow S=\frac{5^{2007}-5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(5S=5^2+5^3+..+5^{2007}=\left(5+5^2+5^3+..+5^{2006}\right)+5^{2007}-5\)
hay \(5S=S+5^{2007}-5\Rightarrow S=\frac{5^{2007}-5}{4}\)
mà
\(S=\left(5+5^4\right)+\left(5^2+5^5\right)+\left(5^3+5^6\right)+\left(5^7+5^{10}\right)..+\left(5^{2001}+5^{2004}\right)+\left(5^{2005}+5^{2006}\right)\)
hay \(S=126.5+126.5^2+126.5^3+126.5^7+...+126.5^{2001}+6.5^{2005}\)
mà rõ ràng \(126.5+126.5^2+126.5^3+126.5^7+...+126.5^{2001}\)chia hết cho 126
còn \(6.5^{2005}\) không chia hết cho 126 nên S không chia hết cho 126.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : S = ( 5 + 54 ) + ( 52 + 55 ) + ( 53 + 56 ) + .... + ( 52003 + 52006 )
= 5( 1 + 53 ) + 52 ( 1 + 53 ) + 53 ( 1 + 53 ) + .... + 52003 ( 1 + 53 )
= 5 ( 1 + 125 ) + 52 ( 1 + 125 ) + 53 ( 1 + 125 ) + .... + 52003 ( 1 + 125 )
= 5.126 + 52 . 126 + 53.126 + ..... + 52003 . 126
= 126 ( 5 + 52 + 53 + .... + 52003 ) ⋮ 126
=> A ⋮ 126 ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=5+5^2+5^3+5^4+...+5^{2006}\)
\(5S=5^2+5^3+5^4+5^5+...+5^{2007}\)
\(5S-S=\left(5^2+5^3+5^4+5^5+...+5^{2007}\right)-\left(5+5^2+5^3+5^4+...+5^{2006}\right)\)
\(4S=5^{2017}-5\)
\(S=\frac{5^{2017}-5}{4}\)
\(S=5+5^2+5^3+5^4+....+5^{2006}\)
\(\Rightarrow5S=5\left(5+5^2+5^3+5^4+.....+5^{2006}\right)\)
\(\Rightarrow5S-S=\left(5^2+5^3+....+5^{2007}\right)-\left(5+5^2+5^3+....+5^{2006}\right)\)
\(\Rightarrow4S=5^{2007}-3\)
\(\Rightarrow S=\frac{5^{2007}-3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì S có 2006 số hạng nên ta chia S thành 334 nhóm mỗi nhóm có 6 số hạng và còn thừa 2 số hạng như sau:
S=5+52+[(53+56)+(54+57)+(55+58)]+.......+[(52001+52004)+(52002+52005)+(52003+52006)]=30+[53(1+125)+54(1+125)+55(1+125)]+.....+[52001(1+125)+52002(1+125)+52003(1+125)]=30+53.126+54.126+55.126+....+52001.126+52002.126+52003.126
=30+126(53+54+55+......+52001+52002+52003)=>S chia 126 dư 30
=> S không chia hết cho 126 (đpcm)
Thank you bn svtkvtm rất nhìu nhé.Mk có đăng thêm mấy câu nữa đấy cậu giúp mk vs
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5S=5^2+5^3+5^4+...+5^{2006}+5^{2007}\)
\(5S-S=\left(5^2+5^3+...+5^{2007}\right)-\left(5+5^2+5^3+...+5^{2006}\right)\)
\(4S=\left(5^{2007}-5\right)\)
\(S=\frac{\left(5^{2007}-5\right)}{4}\)
b)\(S=\left(5+5^4\right)+\left(5^2+5^5\right)+...+\left(5^{2003}+5^{2006}\right)\)
\(S=5.\left(1+5^3\right)+5^2.\left(1+5^3\right)+...+5^{2003}.\left(1+5^3\right)\)
\(S=5.126+5^2.126+...+5^{2003}.126\)
\(S=126.\left(5+5^2+...+5^{2003}\right)\)
vì\(126.\left(5+562+...+5^{2003}\right)\)chia hết cho 126
nên \(S\)chia hết 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