Câu 29: \(A=1+3+3^2+\cdots+3^{100}\)
=>\(3A=3+3^2+3^3+\cdots+3^{101}\)
=>3A-A=\(3+3^2+3^3+\cdots+3^{101}-1-3-3^2-\cdots-3^{100}\)
=>\(2A=3^{101}-1\)
=>\(2A+1=3^{101}\)
=>\(3^{n}=3^{101}\)
=>n=101
Câu 28: B
Câu 27: \(G=\frac35+\frac{3}{5^4}+\frac{3}{5^7}+\cdots+\frac{3}{5^{100}}\)
=>\(5^3\cdot G=3\cdot5^2+\frac35+\frac{3}{5^4}+\cdots+\frac{3}{5^{97}}\)
=>125G-G=\(3\cdot5^2+\frac35+\frac{3}{5^4}+\cdots+\frac{3}{5^{97}}-\frac35-\frac{3}{5^4}-\cdots-\frac{3}{5^{100}}\)
=>124G=\(75-\frac{3}{5^{100}}=\frac{75\cdot5^{100}-3}{5^{100}}\)
=>\(G=\frac{75\cdot5^{100}-3}{124\cdot5^{100}}\)
Đúng 0
Bình luận (0)
