Lời giải:
a)
\(A=1-3+3^2-3^3+3^4-3^5+..+3^{98}-3^{99}\)
\(=(1-3+3^2-3^3)+(3^4-3^5+3^6-3^7)+....+(3^{96}-3^{97}+3^{98}-3^{99})\)
\(=(1-3+3^2-3^3)+3^4(1-3+3^2-3^3)+...+3^{96}(1-3+3^2-3^3)\)
\(=(1-3+3^2-3^3)(1+3^4+...+3^{96})=-20(1+3^4+...+3^{96})\vdots 20\)
Vậy $A$ chia hết cho $20$
b)
\(A=1-3+3^2-3^3+3^4-3^5+...+3^{98}-3^{99}\)
\(3A=3-3^2+3^3-3^4+3^5-3^6+...+3^{99}-3^{100}\)
Cộng theo vế:
\(\Rightarrow A+3A=1-3^{100}\)
\(\Rightarrow A=\frac{1-3^{100}}{4}\)
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