xét \(\frac{a}{n.\left(n+a\right)}=\frac{\left(n+a\right)-n}{n.\left(n+a\right)}=\frac{n+a}{n.\left(n+a\right)}-\frac{n}{n.\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)

vậy ............................

\(\frac{1}{n.\left(n+1\right)}=\frac{\left(n+1\right)-n}{n.\left(n+1\right)}\)

\(=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\)

\(=\frac{1}{n}-\frac{1}{n+1}\left(dpcm\right)\)

Xét \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\left(đpcm\right)\)

b) A=1/2.3+1/3.4+....+1/99.100

=> A=1/2-1/3+1/3-1/4+....+1/99-1/100

=> A=1/2-1/100

=> A=50/100-1/100

=> A=49/100

49/100 

k nhe

49/100

Ta có :

\(1-\frac{3}{n\left(n+2\right)}=\frac{n^2+2n-3}{n\left(n+2\right)}=\frac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(\Rightarrow A=\frac{1.5}{2.4}.\frac{2.6}{3.5}...\frac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)

\(=\left(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n-1}{n}\right)\left(\frac{5}{4}.\frac{6}{5}.\frac{7}{6}...\frac{n+3}{n+2}\right)\)

\(=\frac{1}{n}.\frac{n+3}{4}=\frac{n+3}{n}.\frac{1}{4}\ge\frac{1}{4}\left(dpcm\right)\)

\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{n}-\frac{1}{n+3}\)

     \(=1-\frac{1}{n+3}<1\)

Ta có : 

\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)

\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)

\(A< \frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)\)

\(A< \frac{1}{4}-\frac{1}{4n}\)

Lại có \(n>0\) nên \(\frac{1}{4n}>0\)

\(\Rightarrow\)\(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

đăng thể hiện mình giỏi hả nhóc, lô ga rít lớp 9 đã hc à, 

ối giồi ôi lun, lo ga rít lớp mấy cx ko bít, bv: