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a) Ta có: \(S=5+5^2+5^3+...+5^{96}\)
\(\Leftrightarrow S=\left(5+5^4\right)+\left(5^2+5^5\right)+\left(5^3+5^6\right)+...+\left(5^{93}+5^{96}\right)\)
Vì mỗi cặp của đa thức \(S\)có hai hạng tử nên tổng số cặp là: \(\frac{96}{2}=48\)( cặp )
\(\Rightarrow\)Đa thức \(S\)không dư số nào
\(\Leftrightarrow S=\left(5+5^4\right)+\left(5^2+5^5\right)+\left(5^3+5^6\right)+...+\left(5^{93}+5^{96}\right)\)
\(\Leftrightarrow S=5.\left(5^0+5^3\right)+5^2\left(5^0+5^3\right)+5^3.\left(5^0+5^3\right)+...+5^{93}.\left(5^0+5^3\right)\)
\(\Leftrightarrow S=5.126+5^2.126+5^3.126+...+5^{93}.126\)
\(\Leftrightarrow S=\left(5+5^2+5^3+...+5^{93}\right).126⋮126\)
Vậy \(S⋮126\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(S=5+5^2+5^3+...+5^{96}\)
\(S=\left(5+5^2+5^3+5^4+5^5+5^6\right)+...+\left(5^{91}+5^{92}+5^{93}+5^{94}+5^{95}+5^{96}\right)\)
\(S=5.\left(1+5+5^2+5^3+5^4+5^5\right)+...+5^{91}.\left(1+5^2+5^3+5^4+5^5\right)\)
\(S=5.3906+...+5^{91}.3906\)
\(S=3906.\left(5+...+5^{96}\right)\)
\(S=3.126.\left(5+...+5^{91}\right)\) chia hết cho \(6.\)
b) Do \(S\) là tổng các lũy thừa có cơ số là \(5\).
Cho nên mỗi lũy thừa đều tận cùng là \(5\).
Mà \(S\) có tất cả \(96\) số
\(\Rightarrow\) Chữ số tận cùng của \(S\) là \(0\).
\(S=5+5^2+5^3+..+5^{96}\)
\(S=\left(5+5^2+5^3+5^4+5^5+5^6\right)+\left(5^7+5^8+5^9+5^{10}+5^{11}+5^{12}\right)+...+\left(5^{91}+5^{92}+5^{93}+5^{94}+5^{95}+5^{96}\right)\)\(S=1\left(5+5^2+5^3+5^4+5^6\right)5^6\left(5+5^2+5^3+5^4+5^5+5^6\right)+...+5^{90}+\left(5+5^2+5^3+5^4+5^5+5^6\right)\)\(S=\left(5+5^2+5^3+5^4+5^5+5^6\right)\left(1+5^6+...+5^{90}\right)\)\(S=19530\left(1+5^6+...+5^{90}\right)\)
\(S=155.126.\left(1+5^6+...+5^{90}\right)\)
\(S⋮126\rightarrowđpcm\)
\(S=5+5^2+5^3+...+5^{96}\)
\(S=\overline{...5}+\overline{...5}+\overline{...5}+\overline{...5}+...+\overline{...5}+\overline{...5}\)\(S=\left(\overline{...5}+\overline{...5}\right)+\left(\overline{...5}+\overline{...5}\right)+...+\left(\overline{...5}+\overline{...5}\right)\)\(S=\overline{...0}+\overline{...0}+\overline{...0}\)
\(S=\overline{...0}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
Ta có:
S = 5 + 52 + 53 + ... + 596
S = ( 5 + 54 ) + ( 52 + 55 ) + ( 53 + 56 ) + ... + ( 592 + 595 ) + ( 593 + 596 )
S = 5( 1 + 53 ) + 52( 1 + 53 ) + 53( 1 + 53 ) + ... + 592( 1 + 53 ) + 593( 1 + 53 )
S = 5( 1 + 125 ) + 52( 1 + 125 ) + 53( 1 + 125 ) + ... + 592( 1 + 125 ) + 593( 1 + 125 )
S = ( 1 + 125 )( 5 + 52 + 53 + ... + 592 + 593 )
S = 126( 5 + 52 + 53 + ... + 592 + 593 )
Mà \(126⋮126\)
=> \(126\left(5+5^2+5^3+...+5^{92}+5^{93}\right)⋮126\)
Vậy \(S=5+5^2+5^3+...+5^{96}⋮126\)
# Học tốt #
![](https://rs.olm.vn/images/avt/0.png?1311)
S=(5+5^2+5^3)+(5^4+5^5+5^6)+...+(5^28+5^29+5^30)
=>Có 30:3=10 nhóm
=>S=5(1+5+5^2)+...+5^28(1+5+5^2)
=>S=5.31+...+5^28.31
S=31(5+....+5^28) chia hết cho 31
nhớ bấm đúng cho mình bạn nhé
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Bài 3:
a: a*S=a^2+a^3+...+a^2023
=>(a-1)*S=a^2023-a
=>\(S=\dfrac{a^{2023}-a}{a-1}\)
b: a*B=a^2-a^3+...-a^2023
=>(a+1)B=a-a^2023
=>\(B=\dfrac{a-a^{2023}}{a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(3+3^{3+3^3}\right)+.....+\left(3^{97}+3^{98}+3^{99}\right)\)
\(S=39.1+39.3^3+....+39.3^{96}=>S=39\left(1+3^3+3^6+.....+3^{96}\right)\)
Vậy S chia hết cho 39
a, S = 5+52 +53+....+52022
=> 5S = 52 +53 +54 +....+52023
=> 5S-S=52023 -5
=> S = \(\dfrac{5^{2023}-5}{4}\)
b,Ta có
S = 5+52 +53+....+52022 (2022 số hạng)
S = (5+52 ) + (53+54) +....+(52021+52022) (1011 cặp)
S = 5(1+5) + 53(1+5) + ...+52021(1+5)
S = 5.6 +53.6+...+52021.6
S = 6(5+53+...+52021)\(⋮\) 6(ĐPCM)
a)
\(S=5+5^2+5^3+...+5^{2022}\\ \Rightarrow5S=5^2+5^3+5^4+...+5^{2023}\\ \Rightarrow5S-S=\left(5^2+5^3+5^4+...+5^{2023}\right)-\left(5+5^2+5^3+...+5^{2022}\right)\\ \Rightarrow4S=5^{2023}-5\\ \Rightarrow S=\dfrac{5^{2023}-5}{4}\)
b)
\(S=5+5^2+5^3+...+5^{2022}\\ =\left(5+5^4\right)+\left(5^2+5^5\right)+...+\left(5^{2019}+5^{2022}\right)\\ =5.\left(1+5^3\right)+5^2.\left(1+5^3\right)+...+5^{2019}.\left(1+5^3\right)\\ =5.126+5^2.126+...+5^{2019}.126\\ =126.\left(5+5^2+...+5^{2019}\right)⋮126\\ \)
Vậy S ⋮ 126