Đáp án là:

Đáp án là:

Đáp án là:

Đáp án là: 1-c, 2-a, 3-e, 4-b, 5-f, 6-d.

1: \(M=1+2+2^2+\cdots+2^{100}\)

=>\(2A=2+2^2+2^3+\cdots+2^{101}\)

=>2A-A=\(2+2^2+2^3+\cdots+2^{101}-1-2-2^2-\cdots-2^{100}\)

=>\(A=2^{101}-1\)

=>Chọn C

2: \(N=1+3+3^2+\cdots+3^{100}\)

=>\(3N=3+3^2+3^3+\cdots+3^{101}\)

=>3N-N=\(3+3^2+\cdots+3^{101}-1-3-\cdots-3^{100}\)

=>2N=\(3^{101}-1\)

=>\(N=\frac{3^{101}-1}{2}\)

=>Chọn E

3: \(P=\frac13+\left(\frac13\right)^2+\cdots+\left(\frac13\right)^{100}\)

=>\(P=\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{100}}\)

=>3P=\(1+\frac13+\cdots+\frac{1}{3^{99}}\)

=>3P-P=\(1+\frac13+\cdots+\frac{1}{3^{99}}-\frac13-\frac{1}{3^2}-\cdots-\frac{1}{3^{100}}\)

=>2P=\(1-\frac{1}{3^{100}}=\frac{3^{100}-1}{3^{100}}\)

=>\(P=\frac{3^{100}-1}{2\cdot3^{100}}\)

=>Chọn G

4: \(Q=\frac13-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+\cdots-\frac{1}{3^{100}}\)

=>3Q=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}\)

=>3Q+Q=\(1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{99}}+\frac13-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+\cdots-\frac{1}{3^{100}}\)

=>4Q=\(1-\frac{1}{3^{100}}=\frac{3^{100}-1}{3^{100}}\)

=>\(Q=\frac{3^{100}-1}{4\cdot3^{100}}\)

=>Chọn A

5: \(R=\left(-\frac12\right)+\left(-\frac12\right)^2+\cdots+\left(-\frac12\right)^{100}\)

=>\(R=-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots+\frac{1}{2^{100}}\)

=>2R=\(-1+\frac12-\frac{1}{2^2}+\cdots+\frac{1}{2^{99}}\)

=>2R+R=\(-1+\frac12-\frac{1}{2^2}+\cdots+\frac{1}{2^{99}}-\frac12+\frac{1}{2^2}-\frac{1}{2^3}+\cdots+\frac{1}{2^{100}}\)

=>3R=\(-1+\frac{1}{2^{100}}=\frac{-2^{100}+1}{2^{100}}\)

=>\(R=\frac{-2^{100}+1}{3\cdot2^{100}}\)

=>Chọn D

Đáp án: