\(D=11...122...5\) (n+1 số 2, n số 1).
\(=5+2\left(10+10^2+10^3+...+10^{n+1}\right)+\left(10^{n+2}+10^{n+3}+10^{n+4}+...+10^{2n+1}\right)\)
\(=5+2.10\left(1+10+10^2+...+10^n\right)+10^{n+2}\left(1+10+10^2+...+10^{n-1}\right)\)
\(=5+2.10.\dfrac{10^{n+1}-1}{9}+10^{n+2}.\dfrac{10^n-1}{9}\)
\(=5+\dfrac{20.10^{n+1}-20}{9}+\dfrac{10^{2n+2}-10^{n+1}.10}{9}\)
\(=\dfrac{10^{2n+2}-10^{n+1}.10+20.10^{n+1}-20+45}{9}\)
\(=\dfrac{10^{2n+2}+10.10^{n+1}+25}{9}\)
\(=\left(\dfrac{10^{n+1}+5}{3}\right)^2\).
- Vì \(10^{n+1}\) chia 3 dư 1, 5 chia 3 dư 2.
\(\Rightarrow\left(10^{n+1}+5\right)⋮3\) nên \(\dfrac{10^{n+1}+5}{3}\) là số tự nhiên.
- Vậy D là số chính phương.
