Question

[Rin Itoshi]

Answer 1

[Kẻ Mạo Danh]

a) \(M=1+2+2^2+2^3+\cdots+2^{100}\)

\(2M=2+2^2+2^3+2^4+\cdots+2^{101}\)

\(2M-M=\left(2+2^2+2^3+2^4+\cdots+2^{101}\right)-\left(1+2+2^2+2^3+\cdots+2^{100}\right)\)

\(\Rightarrow M=2^{101}-1\)

Vậy \(M=2^{101}-1\)

b) \(N=1+3^2+3^4+3^6+\cdots+3^{100}\)

\(3N=3+3^2+3^4+3^6+3^8+\cdots+3^{102}\)

\(3N-N=\left(3+3^2+3^4+3^6+3^8+\cdots+3^{102}\right)-\left(1+3+3^2+3^4+3^6+\cdots+3^{100}\right)\)

\(\Rightarrow2N=3^{102}-1\)

\(\Rightarrow N=\frac{3^{102}-1}{2}\)

Vậy \(N=\frac{3^{102}-1}{2}\)

c) \(P=1+5^3+5^6+5^9+\cdots+5^{99}\)

\(5^3\cdot P=5^3+5^6+5^9+5^{12}\cdots+5^{102}\)

\(125P-P=\left(5^3+5^6+5^9+5^{12}\cdots+5^{102}\right)-\left(1+5^3+5^6+5^9+\cdots+5^{99}\right)\)

\(\Rightarrow124P=5^{102}-1\)

\(\Rightarrow P=\frac{5^{102}-1}{124}\)

Vậy \(P=\frac{5^{102}-1}{124}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

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

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

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

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

b: \(N=1+3^2+3^4+\cdots+3^{100}\)

=>\(9N=3^2+3^4+3^6+\cdots+3^{102}\)

=>\(9N-N=3^2+3^4+\cdots+3^{102}-1-3^2-\cdots-3^{100}\)

=>\(8N=3^{102}-1\)

=>\(N=\frac{3^{102}-1}{8}\)

c: \(P=1+5^3+5^6+\cdots+5^{99}\)

=>\(125P=5^3+5^6+5^9+\cdots+5^{102}\)

=>\(125P-P=5^3+5^6+\cdots+5^{102}-1-5^3-\cdots-5^{99}\)

=>\(124P=5^{102}-1\)

=>\(P=\frac{5^{102}-1}{124}\)