Câu 1: 

\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\)

\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\)

\(=\dfrac{5}{4}\cdot\dfrac{4n+3-3}{3\left(4n+3\right)}=\dfrac{5}{4}\cdot\dfrac{4n}{3\left(4n+3\right)}=\dfrac{5n}{3\left(4n+3\right)}\)

Câu 2: 

\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\right)\)

\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right)\)

\(=\dfrac{3}{5}\cdot\dfrac{5n+4-9}{9\left(5n+4\right)}=\dfrac{3}{5}\cdot\dfrac{5\left(n-1\right)}{9\left(5n+4\right)}=\dfrac{n-1}{3\left(5n+4\right)}< \dfrac{1}{15}\)

\(\dfrac{5}{3\cdot7}+\dfrac{5}{7\cdot11}+\dfrac{5}{11\cdot15}+...+\dfrac{5}{\left(4n-1\right)\left(4n+3\right)}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\\ =\dfrac{5}{4}\cdot\dfrac{4n}{12n+9}\\ =\dfrac{5n}{12n+9}\)

Mk thực sự nghĩ đề hình như bị sai hay sao ấy!

2+12345678-5=

a)

ta có:

\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)

\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)

Thay (*) vào dãy A

\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)

\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)

B) tương tự

dễ quá nên tự làm nhé

\(A=\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right).\left(4n+3\right)}\)

\(\frac{4}{5}.A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right).\left(4n+3\right)}\)

\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\)

\(\frac{4}{5}.A=\frac{1}{3}-\frac{1}{4n+3}\)

\(\frac{4}{5}.A=\frac{4n+3}{12n+9}-\frac{3}{12n+9}\)

\(\frac{4}{5}.A=\frac{4n}{12n+9}\)

\(A=\frac{4n}{12n+9}:\frac{4}{5}\)

\(A=\frac{4n}{12n+9}.\frac{5}{4}\)

\(A=\frac{5n}{12n+9}\)

Đề bài sai nha bn

Ủng hộ mk nha ^_^