Question

Chứng minh rằng:

\(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}< \dfrac{3}{16}\)

Answer 1

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

3A=1-2/3+3/3^2-4/3^3+...+99/3^98-100/3^99

3A+A=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99-100/3^100

<1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99

Đặt S=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99

3S=3-1+1/3-1/3^2+1/3^3-...-1/3^98

3S+S=3-1/3^99

S=(3-1/3^99) :4

S=3/4-1/4.3^99

\(\Rightarrow\)4A<3/4-1/4.3^99

\(\Rightarrow\)A<(3/4-1/4.3^99):4

\(\Rightarrow\)A<3/16-1/16.3^99<3/16

Vậy 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16