Question

M = 1+3+3^2+3^3+ .....+3^98+3^99 và N = 3^100/2. Tính N-M

Answer 1

\(M=1+3+3^2+3^3+..+3^{98}+3^{99}\)

\(3M=3+3^2+3^3+3^4+...+3^{99}+3^{100}\)

\(3M-M=2M=3^{100}-1\)

=> \(M=\dfrac{3^{100}-1}{2}\)

\(N-M=\dfrac{3^{100}-3^{100}+1}{2}=\dfrac{1}{2}\)

Answer 2

Giải:

M=1+3+3²+3³+...+3⁹⁸+3⁹⁹

3M=3+3²+3³+3⁴+...+3⁹⁹+3¹⁰⁰

3M-M=(3+3²+3³+3⁴+...+3⁹⁹+3¹⁰⁰)-(1+3+3²+3³+...+3⁹⁸+3⁹⁹)

2M=3¹⁰⁰-1

M=3¹⁰⁰-1/2

⇒N-M

=3¹⁰⁰/2 - (3¹⁰⁰-1)/2

=(3¹⁰⁰-3¹⁰⁰+1)/2

=1/2

Chúc bạn học tốt!