Question

[Thúy Khuất]

cho:M=1+2^4+2^8+...+2^2012+2^2016;N=1+2^2+2^4+2...+2^2016+2^2018.Tính N/M

Answer 1

[Kurosaki Akatsu]

M = 1 + 2⁴ + 2⁸ + ............. + 2²⁰¹² + 2²⁰¹⁶

16M = 2⁴ + 2⁸ + ............. + 2²⁰¹² + 2²⁰¹⁶ + 2²⁰²⁰

16M - M = (2⁴ + 2⁸ + ............. + 2²⁰¹² + 2²⁰¹⁶ + 2²⁰²⁰) - ( 1 + 2⁴ + 2⁸ + ............. + 2²⁰¹² + 2²⁰¹⁶)

15M = 2²⁰²⁰ - 1

M = \(\frac{2^{2020}-1}{15}\)

N = 1 + 2² + 2⁴ + ............. + 2²⁰¹⁶ + 2²⁰¹⁸

4N = 2² + 2⁴ + ............. + 2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰

4N - N = (2² + 2⁴ + ............. + 2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰) - ( 1 + 2² + 2⁴ + ............. + 2²⁰¹⁶ + 2²⁰¹⁸)

3N = 2²⁰²⁰ - 1

N = \(\frac{2^{2020}-1}{3}\)

\(\frac{N}{M}=\frac{2^{2020}-1}{3}:\frac{2^{2020}-1}{15}=\frac{2^{2020}-1}{3}.\frac{15}{2^{2020}-1}=\frac{15}{3}=5\)