Question

Chứng minh rằng 3/4 + 8/9 + 15/16 + ... + 9999/10000 > 98

Đây là đề khảo sát giữa kỳ của mk nhưng mk k bt lm !!

Answer 1

Đặt \(A=\frac{3}{4}+\frac{8}{9}+..........+\frac{9999}{10000}\)

\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+..........+\left(1-\frac{1}{10000}\right)\)

\(=1-\frac{1}{2^2}+1-\frac{1}{3^2}+...........+1-\frac{1}{100^2}\)

\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{100^2}\right)\)\(>99-\left(\frac{1}{1.2}+\frac{1}{2.3}+..........+\frac{1}{99.100}\right)\)

\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{99}-\frac{1}{100}\right)\)

\(=99-\left(1-\frac{1}{100}\right)=99-1+\frac{1}{100}=98+\frac{1}{100}>98\)

Answer 2

=1-1/4+1-1/9+1-1/16+...+1-1/10000

=(1+1+1+...+1)+(-1/4-1/9-1/16-...-1/10000)

=99+(-1/4-1/9-1/16-...-1/10000)

Vì 99+(-1/4-1/9-1/16-...-1/10000)>98

=>3/4 + 8/9 + 15/16 + ... + 9999/10000>98

Vây 3/4 + 8/9 + 15/16 + ... + 9999/10000 >98 

Answer 3

=1-1/4+1-1/9+1-1/16+...+1-1/10000

=(1+1+1+...+1)+(-1/4-1/9-1/16-...-1/10000)

=99+(-1/4-1/9-1/16-...-1/10000)

Vì 99+(-1/4-1/9-1/16-...-1/10000)>98

=>3/4 + 8/9 + 15/16 + ... + 9999/10000>98

Vây 3/4 + 8/9 + 15/16 + ... + 9999/10000 >98 

Answer 4

1/4 = 1/2.2 < 1/1.2 = 1 - 1/2 
1/9 = 1/3.3 < 1/2.3 = 1/2 - 1/3 
1/16 = 1/4.4 < 1/3.4 = 1/3 - 1/4 
.......................................... 
1/10000 = 1/100.100 < 1/99.100 = 1/99 - 1/100 
=> N = 1/4 + 1/9 + 1/16 + ... + 1/10000 < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/99 - 1/100 = 1 - 1/100 < 1 
M = 3/4 + 8/9 + 15/16 +...+ 9999/10000 = (1 - 1/4) + (1 - 1/9) + (1 - 1/16) + ...+ (1 - 1/10000) = 99 - (1/4 + 1/9 + 1/16 +...+ 1/10000) = 99 - N 
=> 98 < M