Ta có

\(M=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^3.3^2}+.....+\frac{100^2-99^2}{99^2.100^2}\)

\(M=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+......+\frac{1}{99^2}-\frac{1}{100^2}\)

\(M=1-\frac{1}{100^2}< 1\)

=> M<1

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)

 \(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)

\(\text{ta thấy }\frac{3}{1^2.2^2}=\frac{1}{1^2}-\frac{1}{2^2};\frac{5}{2^2.3^2}=\frac{1}{2^2}-\frac{1}{3^2};....;\frac{199}{99^2.100^2}=\frac{1}{99^2}-\frac{1}{100^2}\)

\(\text{suy ra }\)\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{199}{99^2.100^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)

\(=\frac{1}{1^2}-\frac{1}{100^2}=\frac{1}{1}-\frac{1}{10000}=\frac{10000}{10000}-\frac{1}{10000}=\frac{9999}{10000}

\(\Rightarrow\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{199}{9801.1000}\)

\(\Rightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{9801}-\frac{1}{1000}\)

\(\Rightarrow1-\frac{1}{1000}< 1\)

\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)

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

\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)

\(T=\frac{1}{\left(2-1\right)\left(2+1\right)}+\frac{1}{\left(3-1\right)\left(3+1\right)}+...+\frac{1}{\left(50-1\right)\left(50+1\right)}\)

\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)

Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T

JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)

a)Xét vế trái , ta có :

Gọi tổng các số hạng ở vế trái là A

=> A= \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\)

=>3A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)

=> 3A - A = 1 + \(\frac{1}{3}\)+ \(\frac{1}{3^2}\)+ ... + \(\frac{1}{3^{98}}\)- ( \(\frac{1}{3}\)+\(\frac{1}{3^2}\)+ ... +\(\frac{1}{3^{99}}\))

=> 2A = 1 - \(\frac{1}{3^{99}}\)

=> A = \(\frac{1}{2}\)- \(\frac{1}{3^{99}.2}\) < \(\frac{1}{2}\)

b)\(\frac{3}{1^2.2^2}\)+ \(\frac{5}{2^2.3^2}\)+ ... + \(\frac{19}{9^2.10^2}\)

= \(\frac{3}{1.4}\)+ \(\frac{5}{4.9}\)+ .... + \(\frac{19}{81.100}\)

= 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{9}\)+ ... + \(\frac{1}{81}\)- \(\frac{1}{100}\)

= 1 - \(\frac{1}{100}\) <1

a,

\(\sum\limits^{99}_{x=1}\left(\frac{1}{3^x}\right)=\frac{1}{2}\)

bài a nó có ............

ta có:

 \(\frac{1}{11}\)>\(\frac{10}{20}\)

\(\frac{1}{12}\)>\(\frac{10}{20}\)

\(\frac{1}{13}\)>\(\frac{10}{20}\)

....

\(\frac{1}{19}\)>\(\frac{10}{20}\)

=>E >\(\frac{10}{20}\)

vậy E > \(\frac{1}{2}\)