$\frac{4}{n\left(n+2\right)\left(n+4\right)}=\frac{n+4-n}{n\left(n+2\right)\left(n+4\right)}=\frac{1}{n\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+4\right)}$4n(n+2)(n+4) =n+4−nn(n+2)(n+4) =1n(n+2) −1(n+2)(n+4) $\frac{B}{9}=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}=\frac{1}{3}-\frac{1}{27.29}<\frac{1}{3}$B9 =11.3 −13.5 +13.5 −15.7 +...+125.27 −127.29 =13 −127.29 <13 $\Rightarrow B<3$

Ta có :

\(B=\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+...............+\dfrac{36}{25.27.29}\)

\(B=9\left(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+\dfrac{4}{5.7.9}+.............+\dfrac{4}{25.27.29}\right)\)

\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}+\dfrac{1}{5.7}-\dfrac{1}{7.9}+...........+\dfrac{1}{25.27}-\dfrac{1}{27.29}\right)\)

\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{27.29}\right)\)

\(B=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\)

\(B=9.\dfrac{1}{3}-9.\dfrac{1}{783}\)

\(B=3-\dfrac{9}{783}< 3\)

\(\Rightarrow B< 3\rightarrowđpcm\)

\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)

Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)

2.A=2.(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\))

2. A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+\(\dfrac{2}{3.4.5}\)+...+\(\dfrac{2}{17.18.19}\)

2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+ ...+\(\dfrac{1}{17.18}\)-\(\dfrac{1}{18.19}\)

2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{18.19}\)=\(\dfrac{85}{171}\)

A=\(\dfrac{85}{171}\):2=\(\dfrac{85}{342}\)

Ta cũng có: \(\dfrac{1}{4}\) = \(\dfrac{171}{684}\); \(\dfrac{85}{342}\) = \(\dfrac{170}{684}\)

Vì 170 < 171 ( \(\dfrac{170}{684}\) < \(\dfrac{171}{684}\) )

Vậy \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\) Gio thi tu ma lam ko thích viết nữa mệt

A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}\)

Theo công thức:

\(\dfrac{2m}{b.\left(b+m\right).\left(b+2m\right)}=\dfrac{1}{b.\left(b+m\right)}-\dfrac{1}{\left(b+m\right).\left(b+m.2\right)}\)Ta có:

2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{18.19.20}\)

2A=\(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)2A=\(\dfrac{1}{1.2}-\dfrac{1}{19.20}\)

2A=\(\dfrac{1}{2}-\dfrac{1}{19.20}\)

A=\(\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)

A=\(\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\)

A=\(\dfrac{1}{2}.\dfrac{19.20-2}{2.19.20}\)

A=\(\dfrac{19.20-2}{2.2.19.20}\) < \(\dfrac{19.20}{2.2.19.20}\) = \(\dfrac{1}{4}\)

\(\Rightarrow\) A<\(\dfrac{1}{4}\)

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

\(A=\dfrac{1}{1}-\dfrac{1}{20}\)

\(A=\dfrac{20}{20}-\dfrac{1}{20}\)

\(A=\dfrac{19}{20}\)

A = \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{18.19.20}\)

A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{18.19}\)-\(\dfrac{1}{19.20}\)

A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{19.20}\)

A = \(\dfrac{1}{2}\)-\(\dfrac{1}{380}\)

A = \(\dfrac{189}{380}\)

(Mình nghĩ là vậy, có gì sai bạn bỏ qua nha )

A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)

A=\(\dfrac{1}{1}-\dfrac{1}{39}\)

A=\(\dfrac{38}{39}\)

còn lại tự làm do mình có việc chút

Chưa học

Bài này mình da làm roi dễ

S = \(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+......+\dfrac{1}{10000}\)

\(\Rightarrow S=\dfrac{1}{4.1}+\dfrac{1}{4.4}+\dfrac{1}{4.9}+.....+\dfrac{1}{4.2500}\)

\(\Rightarrow S=\dfrac{1}{4.\left(1+\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}\right)}< \dfrac{1}{2}\)

\(\RightarrowĐPCM\)