Question

Bài 6. Cho \( S = \frac{1}{3} - \frac{2}{3^2} + \frac{3}{3^3} - \frac{4}{3^4} + \ldots + \frac{99}{3^{99}} - \frac{100}{3^{100}} \). So sánh \( S \) và \( \frac{1}{5} \).

Answer 1

Ta có:

A=1/3 - 2/3^2+3/3^3 - 4/3^4+ ... - 100/3^100

=3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99

=4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100

=12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99

=16A=12A+4A=3-1/3^99-100/3^99-100/3^1...

=16A=3-101/3^99-100/3^100

=A=3/16-(101/3^99+100/3^100)/16 < 3/16

Ta suy ra :A<3/16