Ta thấy:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}

\(A = \dfrac{1}{2^2} + \dfrac{1}{4^2} +\dfrac{1}{6^2} +...... +\dfrac{1}{100^2} \)

\(A = \dfrac{1}{1^2.2^2} +\dfrac{1}{2^2.2^2} +\dfrac{1}{2^2.3^2} + .......+\dfrac{1}{2^2.2^{50}}\)

\(A = \dfrac{1}{2^2}.(\) \( \dfrac{1}{1^2} + \dfrac{1}{2^2} +\dfrac{1}{3^2} +...... +\dfrac{1}{50^2}) \)

\(A < \dfrac{1}{2^2}.( \dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{49.50}\) \()\)

\(= \dfrac{1}{2^2}.(1-\dfrac{1}{2} + \dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{49}-\dfrac{1}{50})\)

\(= \dfrac{1}{2^2} . ( 1 - \dfrac{1}{50})\)

\(< \dfrac{1}{2^2} . 2 = \dfrac{1}{2}\)

TA CÓ 1/2^2=1/4

1/3^2<1/2.3=1/2-1/3

1/4^2<1/3.4=1/3-1/4

1/100^2<1/99.100

=>1/2^2+2/3^2+.....+1/100^2<1/1.2+1/2.3+..+1/99.100

=1-99/100=99/100<1

! là j z

\("!"\)  là giai thừa đó bạn ạ .

\(VD:\)  \(3!=1.2.3=6\)

          \(4!=1.2.3.4=24\)

ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều