Ta có:
A= $1$ + 3/2^3+4/2^4+....+100/2^100
2.A= $2$ + 3/2^2+ 4/2^3+...+ 100/ 2^99
2.A - A= ( $2$ + 3/2^2+ 4/2^3+...+ 100/ 2^99) - ( $1$ + 3/2^3+4/2^4+....+100/2^100)
A= ( 2- 1) + ( 3/4- 100/ 2^ 100) + ( 4/2^ 3- 3/2^3) + ( 5/2^4- 4/2^ 4)+...+ ( 100/ 2^99- 99/2^ 99)
A= 1+ ( 3/4- 100/ 2^ 100) + 1/2^3+ 1/2^4+...+ 1/2^ 99
Đặt B= 1/2^3+ 1/2^4+...+ 1/2^ 99
A= 1+ 3/4+ 100/ 2^ 100+ B
B= 1/2^3+ 1/2^4+...+ 1/2^ 99
2.B= 1/2^ 2+ 1/2^3+...+ 1/2^98
2.B- B=( 1/2^ 2+ 1/2^ 3+..+ 1/2^98)- (1/2^3+ 1/2^4+...+ 1/2^ 99)
B= 1/2^2- 1/2^ 99
B= 1/4- 1/2^ 99
⇒ A = 1+ 3/4- 100/ 2^ 100+ B
A= 1+ 3/4- 100/ 2^ 100+ 1/4- 1/2^ 99
= 1+ 3/4+ 1/4- ( 100/2^ 100+ 1/2^ 99)
= 1+ 1- ( 100/2^100+ 2/2^100)
= 2- 102/2^ 100
Vậy A= 2- 102/2^ 100
Chúc bạn học tốt!
