Question

[Nguyễn Minh Quân]

1. Chứng minh rằng với n là stn khác 0 thì \(4^{2n+1}+3^{n+2}\)chia hết cho 13.

2.Tính: 

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

Answer 1

[Monkey D.Luffy]

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

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

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

Answer 2

[Trịnh Tiến Đức]

1)

4²ⁿ⁺¹+3ⁿ⁺²= (4²)ⁿ.4 +3ⁿ.3²

                = 16ⁿ.4+3ⁿ.9

               =13ⁿ.4+3ⁿ.4+3ⁿ.9

              =13ⁿ.4+3ⁿ.(4+9)

             = 13ⁿ.4+3ⁿ.13 = 13.(13^n-1+3ⁿ) chia het cho 13

=> 4²ⁿ⁺¹+3ⁿ⁺² chia hết cho 13

2)

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

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

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