Question

[Nguyễn Ngọc Ánh]

Cho dãy số: \(a_{n+2}=2.a_{n+1}-a_n+1\) (\(x\in\) N*). Biết \(a_1=1;a_2=3\)

Chứng minh rằng: \(4.a_n.a_{n+2}+1\) là số chính phương

Answer 1

[Nguyễn Việt Lâm]

Đây là toán 8 thật à :(((((

\(a_{n+2}-a_{n+1}=a_{n+1}-a_n+1\)

Đặt \(b_n=a_{n+1}-a_n\Rightarrow b_{n+1}=a_{n+2}-a_{n+1}\)

\(\Rightarrow b_{n+1}=a_{n+1}-a_n+1=b_n+1\)

Lại có \(b_1=a_{1+1}-a_1=a_2-a_1=2\)

\(\Rightarrow b_2=b_1+1\)

\(\Rightarrow b_3=b_2+1\)

...

\(\Rightarrow b_n=b_{n-1}+1\)

Cộng vế với vế:

\(b_2+b_3+...+b_{n-1}+b_n=b_1+b_2+...+b_{n-1}+1+1+...+1\) (n-1 số 1)

\(\Rightarrow b_n=b_1+1\left(n-1\right)=n+1\)

\(\Rightarrow a_{n+1}-a_n=n+1\)

Từ đó \(\Rightarrow a_{n+1}=a_n+n+1\)

\(\Rightarrow a_n=a_{n-1}+n\)

\(\Rightarrow a_{n-1}=a_{n-2}+n-1\)

...

\(\Rightarrow a_3=a_2+3\)

\(\Rightarrow a_2=a_1+2\)

Lại cộng vế với nhau:

\(a_{n+1}+a_n+...+a_3+a_2=a_n+a_{n-1}+...+a_2+a_1+\left(n+1\right)+n+...+2\)

\(\Rightarrow a_{n+1}=a_1+\left(n+1\right)+n+...+2\)

\(\Rightarrow a_{n+1}=\left(n+1\right)+n+...+2+1\)

\(\Rightarrow a_n=n+n-1+...+1=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow a_{n+2}=\frac{\left(n+2\right)\left(n+3\right)}{2}\)

\(\Rightarrow4a_{n+2}a_n+1=n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1\)

\(=n\left(n+3\right)\left(n+1\right)\left(n+2\right)+1\)

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

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

\(=\left(n^2+3n+1\right)^2\) (đpcm)