Question

Cho \(M=\frac{1+3+3^2+.....+3^{100}}{1+3+3^2+.....+3^{99}}\)    Và           \(N=\frac{1+5+5^2+......+5^{100}}{1+5+5^2+......+5^{99}}\)

So sánh M và N

Answer 1

M=1+  3^100/1+3+3^2+..+3^99

=1+1:   1+3+3^2+...+3^99/3^100

=1+1:(1/3^100+1/3^99+..+1/3)

tương tự ta có

N=1+1:         (1/5^100+1/5^99+......+1/5)

do 1/5^100<1/3^100;1/5^99<1/3^99,...,1/5<1/3

=M<N

Answer 2

M=1+  3^100/1+3+3^2+..+3^99

=1+1:   1+3+3^2+...+3^99/3^100

=1+1:(1/3^100+1/3^99+..+1/3)

tương tự ta có

N=1+1:         (1/5^100+1/5^99+......+1/5)

do 1/5^100<1/3^100;1/5^99<1/3^99,...,1/5<1/3

=M<N