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`->` Mình tham khảo ở đây để làm nếu sai thì cho mik xl ạ.

Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+....+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\)

\(2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\\ 2A=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+....+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)}-\dfrac{1}{\left(n-1\right)\cdot n}\)

\(2A=\dfrac{1}{1\cdot2}-\dfrac{1}{\left(n-1\right)\cdot\left(n-2\right)}\)

\(A=\dfrac{1}{4}-\dfrac{1}{\left(n-1\right)\cdot\left(n-2\right)\cdot2}\)

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\cdot\cdot\cdot+\dfrac{2}{\left(n-2\right)\cdot\left(n-1\right)\cdot n}\right)\)

\(=\dfrac{1}{2}\left[\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+\cdot\cdot\cdot+\dfrac{1}{\left(n-2\right)\left(n-1\right)}-\dfrac{1}{\left(n-1\right)n}\right]\)

\(=\dfrac{1}{2}\left[\dfrac{1}{1\cdot2}-\dfrac{1}{\left(n-1\right)n}\right]\)

\(=\dfrac{1}{2}\cdot\left[\dfrac{n\left(n-1\right)}{2n\left(n-1\right)}-\dfrac{2}{2n\left(n-1\right)}\right]\)

\(=\dfrac{1}{2}\cdot\dfrac{n\left(n-1\right)-2}{2n\left(n-1\right)}\)

\(=\dfrac{n^2-n-2}{4n\left(n-1\right)}\)

#\(Toru\)

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)

\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

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

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

Ta có:

 \(1+2+4+8+...+512=1+2+2^2+2^3+...+2^9\)

Đặt \(A=1+2+2^2+2^3+...+2^9\)

\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^9+2^{10}\)

\(\Rightarrow2A-A=2^{10}-1\)

\(\Rightarrow A=2^{10}-1=1024-1=1023\)

bài này dễ ợt , đợi mình đi họcvề rồi giải cho

\(B=2B-B=2+2^2+2^3+2^4+...+2^{11}-\left(1+2^2+2^3+...+2^{10}\right)=2^{11}-1=2048-1=2047\)

kết bạn với mk đi

2S = 2 + 2 mũ 2 + 2 mũ 3 + ...+ 2 mũ 10 +  2 mũ 11 

2S - S = 2 mũ 11 - 1 

S        = 2 mũ 11 - 1  = 2048 - 1 

S        = 2047 

what your name anh where shool you study

a,Ta có :2A=2+2^2+2^3+...+2^2011

2A-A=2^2011-2^0=2^2011-1

b,Tính 4B làm tương tự A

a)2A=2(1+2.2+2.2²+...+2.2²⁰¹⁰)

=2.1+2.2+2.2²+...+2.2²⁰¹⁰

=2+2²+2³+...+2²⁰¹¹

2A-A=(2+2²+2³+...+2²⁰¹¹)-(1+2+2²+...+2²⁰¹⁰)

A=2²⁰¹⁰-1