a) \(S=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5S=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5S-S=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(\Rightarrow4S=5^{101}-5\)
\(\Rightarrow S=\frac{5^{101}-5}{4}\)
b) \(4S+5=5^x\)
\(\Rightarrow5^{101}-5+5=5^x\)
\(\Rightarrow5^{101}=5^x\)
\(\Rightarrow x=101\)
Vậy x = 101
c) \(S=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(\Rightarrow S=\left(5+25\right)+5^2.\left(5+5^2\right)+...+5^{98}.\left(5+5^2\right)\)
\(\Rightarrow S=30+5^2.30+...+5^{98}.30\)
\(\Rightarrow S=\left(1+5^2+...+5^{98}\right).30⋮30\)
\(\Rightarrow S⋮30\left(đpcm\right)\)
a)\(S=5+5^2+...+5^{100}\)
\(5S=5\left(5+5^2+...+5^{100}\right)\)
\(5S=5^2+5^3+...+5^{101}\)
\(5S-S=\left(5^2+5^3+...+5^{101}\right)-\left(5+5^2++...+5^{100}\right)\)
\(4S=5^{101}-5\)
\(S=\frac{5^{101}-5}{4}\)
b)Theo câu a ta có:
\(4S+5=5^x\Leftrightarrow5^{101}-5+5=5^x\)
\(\Leftrightarrow5^{101}=5^x\Leftrightarrow x=101\)
c)\(S=5+5^2+...+5^{100}\)
\(=\left(5+5^2+5^3\right)+...+\left(5^{98}+5^{99}+5^{100}\right)\)
\(=5\left(1+5+5^2\right)+...+5^{98}\left(1+5+5^2\right)\)
\(=5\cdot31+...+5^{98}\cdot31\)
\(=31\cdot\left(5+...+5^{98}\right)⋮31\)
a, \(S=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5S=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5S-S=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(\Rightarrow4S=5^{101}-5\)
\(\Rightarrow S=\frac{5^{101}-5}{4}\)
b, Ta có : \(4S+5=5^x\)
\(\Rightarrow5^{101}-5+5=5^x\)
\(\Rightarrow5^{101}=5^x\Rightarrow x=101\)
Vậy x = 101
c, Ta có : Số số hạng của dãy số S là : ( 100 - 1 ) : 1 + 1 = 100 ( số hạng )
Vậy ta có số nhóm là : 100 : 2 = 50 ( nhóm )
\(S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(S=\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{98}\left(5+5^2\right)\)
\(S=1.30+5^2.30+...+5^{98}.30\)
\(S=\left(1+5^2+...+5^{98}\right).30\)
Mà : \(1+5^2+...+5^{98}\in N\Rightarrow S⋮30\)
Vậy : S \(⋮\)30
