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S=5+5^2+5^3+...+5^2004
S=(5+5^2+5^3+5^4)+(5^6+5^7+5^8+5^9)+...+(+5^2001+5^2002+5^2003+5^2004)
S=1(5+5^2+5^3+5^4)+5^5(5+5^2+5^3+5^4)+...+5^2000(5+5^2+5^3+5^4)
S=1*780+5^5*780+...+5^2000*780
S=780(1+5^5+..+5^2000)
vì 780 chia hết cho 65 nên S chia hết cho 65
k mik nha
\(S=5+5^2+5^3+...+5^{1992}\)
\(=5\left(1+5\right)+5^3\left(1+5\right)+...+5^{1991}\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{1991}.6=6\left(5+5^3+...+5^{1991}\right)⋮6\)
1) (5+54)+(52+55)+...........+(52003+52006)= 5(1+53)+52(1+53)+..............+52003(1+53)
= (5+52+..........+52003).126 ->S chia hết cho 126
2, 7+73+................+71997+71999 = 7(1+72)+..............+71997(1+72)
= (7+...............+71997).50-> chia hết cho 5
= 7(1+72+.......+71998) -> chia hết cho 7
-> chia hết cho 35
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
Đặt \(A=5+5^2+5^3+5^4+...+5^{49}+5^{50}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{49}+5^{50}\right)\)
\(=5.\left(1+5\right)+5^3.\left(1+5\right)+...+5^{49}.\left(1+5\right)\)
\(=5.6+5^3.6+...+5^{49}.6\)
\(=6.\left(5+5^3+...+5^{49}\right)⋮6\)
Vậy \(A⋮6\)
Số số hạng của A:
98 - 1 + 1 = 98 (số)
Do 98 ⋮ 2 nên ta có thể nhóm các số hạng của A thành các nhóm mà mỗi nhóm có 2 số hạng như sau:
A = (5 + 5²) + (5³ + 5⁴) + ... + (5⁹⁷ + 5⁹⁸)
= 5.(1 + 5) + 5³.(1 + 5) + ... + 5⁹⁷.(1 + 5)
= 5.6 + 5³.6 + ... + 5⁹⁷.6
= 6.(5 + 5³ + ... + 5⁹⁷) ⋮ 6
Vậy A ⋮ 6
A=(5+5^2)+(5^3+5^4)+...+(5^97+5^98)
A=5(1+5)+5^3(1+5)+...+5^97(1+5)
A=(5.6)+(5^3.6)+...+(5^97.6)
A=6.(5+5^3+...+5^97)
suy ra A⋮6
Suy ra A
a) \(S=1+5+5^2+5^3+...+5^{28}\)
\(S=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{27}+5^{28}\right)\)
\(S=1\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)
\(S=\left(1+5^2+...+5^{27}\right).6⋮3\left(dpcm\right)\)
b) \(S=1+5+5^2+5^3+...+5^{28}\)
\(\Rightarrow5S=5+5^2+5^3+5^4+...+5^{29}\)
\(\Rightarrow5S-S=\left(5+5^2+5^3+5^4+...+5^{29}\right)-\left(1+5+5^2+5^3+...+5^{28}\right)\)
\(\Rightarrow4S=5^{29}-1\)
\(\Rightarrow4S+1=5^{29}-1+1\)
\(\Rightarrow4S=5^{29}=5^n\)
\(\Rightarrow n=29\)
a) \(S=1+5+5^2+5^3+...+5^{28}\)
\(\Rightarrow S=\left(1+5\right)+5^2\left(1+5\right)+...+5^{27}\left(1+5\right)\)
\(\Rightarrow S=6+5^2.6+...+5^{27}.6\)
\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮6\)
\(\Rightarrow S=6\left(1+5^2+...+5^{27}\right)⋮3\)
\(\Rightarrow dpcm\)
b) Bạn xem lại đề
\(S=5+5^2+5^3+5^4+...+5^{2004}\)
\(S=\left(5+5^2+5^3+5^4\right)+...+\left(5^{2001}+5^{2002}+5^{2003}+5^{2004}\right)\)
\(S=5\left(1+5+5^2+5^3\right)+...+5^{2001}\left(1+5+5^2+5^3\right)\)
\(S=\left(1+5+5^2+5^3\right)\left(5+5^5+...+5^{2001}\right)\)
\(S=156\left(5+5^5+...+5^{2001}\right)⋮156\)
\(S=5+5^2+5^3+5^4+...+5^{2004}\)
\(S=\left(5+5^2+5^3+5^4\right)+...+\left(5^{2001}+5^{2002}+5^{2003}+5^{2004}\right)\)
\(S=\left(5+5^2+5^3+5^4\right)\left(1+5^4+...+5^{2000}\right)\)
\(S=780\left(1+5^4+...+5^{2000}\right)⋮65\)