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0\(a.S=1-5+5^2-5^3+...+5^{98}-5^{99}\\ 5S=5-5^2+5^3-5^4+.....+5^{99}-5^{100}\\ 5S+S=\left(5-5^2+5^3-5^4+.....+5^{99}-5^{100}\right)+\left(1-5^{ }+5^2-5^3+.....+5^{98}-5^{99}\right)\\ 6S=1-5^{100}\\ S=\dfrac{1-5^{100}}{6}\\ \)
\(b,S6=1-5^{100}\\ 1-S6=5^{100}\)
=> 5100 chia 6 du 1
Bài 1:
a: \(S=1-5+5^2-5^3+...+5^{98}-5^{99}\)
=>\(5S=5-5^2+5^3-5^4+...+5^{99}-5^{100}\)
=>\(6S=5-5^2+5^3-5^4+...+5^{99}-5^{100}+1-5+5^2-5^3+...+5^{98}-5^{99}\)
=>\(6S=-5^{100}+1\)
=>\(S=\dfrac{-5^{100}+1}{6}\)
b: S=1-5+52-53+...+598-599 là số nguyên
=>\(\dfrac{-5^{100}+1}{6}\in Z\)
=>\(-5^{100}+1⋮6\)
=>\(5^{100}-1⋮6\)
=>\(5^{100}\) chia 6 dư 1
Lời giải:
$C=1+5+5^2+5^4+.....+5^{98}+5^{100}$
$25C=5^2C=5^2+5^3+5^4+5^6+....+5^{100}+5^{102}$
$25C-C=(5^3+5^{102})-(5+1)$
$24C=5^{102}-119$
$C=\frac{5^{102}-119}{24}$
\(A=5^3+5^4+5^5+...+5^{100}\)
\(A=5^3\left(1+5^1+5^2+...+5^{97}\right)\)
\(A=5^3.\dfrac{5^{97+1}-1}{5-1}=\dfrac{5^3}{4}.\left(5^{98}-1\right)\)
\(A=5^2+5^4+5^6+...+5^{100}+5^{102}\\ =5^2.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)\\ =25.\left(1+5^2+5^4+...+5^{98}+5^{100}\right)⋮25\)
\(a,16^{19}=\left(2^4\right)^{19}=2^{76}\\ 8^{25}=\left(2^3\right)^{25}=2^{75}\)
Vì \(2^{76}>2^{75}=>16^{19}>8^{25}\)
b,\(3^{500}=\left(3^5\right)^{100}=243^{100}\)
Vì \(243^{100}>5^{100}=>3^{500}>5^{100}\)
(x + 2) + (x + 4) + (x + 6) +...+ (x + 100) = 5100
x + 2 + x + 4 + x + 6 +... + x + 100 = 5100
50x + (2 + 4 + 6 +... + 100) = 5100
50x + 2550 = 5100
50x = 5100 - 2550 = 2550
x = 2550 : 50 = 51
Vây x = 51
(x+1) + (x+3) + ... + (x+99) = 5100
(x+x+..+x) + (1+3+...+99) = 5100
xét dãy số 1;3;...;99 có:
(99-1) : 2 + 1 = 50 (số)
có: (x.50) + (99+1) . 50 : 2 = 5100
(x.50) + 100 .50 : 2 = 5100
(x.50) + 5000 : 2 = 5100
(x.50) + 2500 = 5100
x.50 = 5100 - 2500
x.50 = 2600
x = 2600 : 50
x = 52
vậy......
A=5^100+6=5^100+5+1=5^101+1 ; 5^100+4=5^100+5-1=5^101-1
B=5^100+7=5^100+5+2=5^101+2 ; 5^100+5=5^101
vs nha