\(B=2+2^2+2^3+...+2^{100}\)
\(\Rightarrow2B=2^2+2^3+2^4+...+2^{101}\)
\(\Rightarrow2B-B=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)
\(\Rightarrow2B-B=2^{101}-2\)
\(A=1+3+3^2+3^3+...+3^{99}\\ \Rightarrow A=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\)
\(\Rightarrow A=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=\left(1+3+3^2+3^3\right)\left(1+3^4+...+9^{96}\right)\)
\(\Rightarrow A=40\left(1+3^4+...+9^{96}\right)⋮40\)
câu 1 đã có người giải
câu 2:
A=1+3+32+33+....+399
A=(1+3+32+33)+...+(396+397+398+399)
A=30.(1+3+32+33)+....+396.(1+3+32+33)
A=30.40+...+396.40
A=(30+....+396).40 chia hết cho 40
vậy A chia hết cho 40
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