Question

Chứng minh: 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +...+ 99/3^99 - 100/3^100 < 3/16

Làm ơn giải giúp mk vs nha, xin chân thành cảm ơn

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
Suy ra A<3/16

Answer 2

ai tk mình đi đang bị âm điểm nè

cảm ơn các bạn nhìu!!!

Answer 3

ai tk mf vs

Answer 4

nguyen thi hoa tk cho mink voi mink vua k cho ban day

Answer 5

sai rùi bạn

gg

Answer 6

đây nha

Answer 7

A<3/16

Answer 8

Lời giải:
Đặt �=13−232+333−....+99399−1003100A=31​−322​+333​−....+39999​−3100100​

3�=1−23+332−.....+99398−1003993A=1−32​+323​−.....+39899​−399100​

⇒4�=�+3�=1−13+132−133+....−1399−1003100⇒4A=A+3A=1−31​+321​−331​+....−3991​−3100100​

12�=3−1+13−132+...−1398−10039912A=3−1+31​−321​+...−3981​−399100​

⇒4�+12�=3−100399−1399−1003100<3⇒4A+12A=3−399100​−3991​−3100100​<3

$\Rightarrow 16A<3$

⇒�<316⇒A<163​