Question

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{n\left(n+1\right)}\) tìm công thức rồi giải giúp em với ạ 

Answer 1

cái nài toán lớp 6 nha em

A=1/1-1/2+1/2-1/3+....+1/n-1/n+1

A=1-1/n+1

A=n/n+1

Answer 2

\(A=\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

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

\(A=1-\frac{1}{n+1}\)

\(A=\frac{n}{n+1}\)

   ~~~~~Hok tốt~~~~~

Answer 3

Cứ theo công thức này để phân tích nhé bn:\(\frac{a}{b-c}=\frac{1}{b}-\frac{1}{c}\)(với b-c=a)

Bn thay vào từng phân số rồi sau đó triệt tiêu đi

Answer 4

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}\)

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

\(A=1-\frac{1}{n+1}\)

\(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)

\(A=\frac{n}{n+1}\)

Answer 5

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

Công thức \(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

\(=\frac{n}{n+1}\)

Answer 6

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

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

\(A=1-\frac{1}{n+1}\)

\(A=\frac{n}{n+1}\)

~~~~~hok tốt~~~~

Answer 7

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

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

\(A=1-\frac{1}{n+1}\)

\(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)

\(A=\frac{n+1-1}{n+1}\)

\(A=\frac{n}{n+1}\)