Sửa đề: \(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<2\)

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{50^2}<\frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{49}-\frac{1}{50}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1-\frac{1}{50}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<2\) (ĐPCM)

 ->  M = (100 – 1).(100 – 2^2). (100 – 3^2)…(100 – 50^2)

M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2) .(100 – 10^2) .(100 – 11^2) …(100 – 50^2)

M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2). (100 – 100) .(100 – 11^2) …(100 – 50^2)

M = (100 – 1).(100 – 2^2). (100 – 3^2)… (100 – 9^2) .0.(100 – 11^2) …(100 – 50^2)

M = 0

Vậy M = 0.

Câu a:
Ta có: 1/51 > 1/100 ; 1/52>1/100 ..... ; 1/99>1/100
        => 1/51+1/52+...+1/100 > 1/100+1/100+.....+1/100 ( 50 số ) = 50/100=1/2 (1)
Ta lại có: 1/52<1/51; 1/53<1/51;....; 1/100<1/51
        => 1/51+1/52+....+1/100<1/51+1/51+.......+1/51 ( 50 số = 50/51<1 (2)
  Từ (1) (2) => đpcm
Câu b làm tương tự :)