Question

Bài 1:

a,với mọi số nguyên dương n thì:

\(3^{n+2}-2^{n+2}-2^n\) chia hết cho 10

b, Cho A= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....................+\frac{1}{2007}+\frac{1}{2008}\)

B= \(\frac{2007}{1}+\frac{2006}{2}+\frac{2005}{3}+............+\frac{2}{2006}+\frac{1}{2007}\)

Tính \(\frac{B}{A}\)

Answer 1

Bài 1:

a) Sửa lại là: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\) nhé.

\(3^{n+2}-2^{n+2}+3^n-2^n\)

\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)

\(=3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)

\(=3^n.\left(9+1\right)-2^n.\left(4+1\right)\)

\(=3^n.\left(9+1\right)-2^{n-1}.2.\left(4+1\right)\)

\(=3^n.10-2^{n-1}.2.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10.\left(3^n-2^{n-1}\right)\)

Vì \(10⋮10\) nên \(10.\left(3^n-2^{n-1}\right)⋮10.\)

\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\left(đpcm\right)\left(\forall n\in N^X\right).\)

Chúc bạn học tốt!