\(C=44...488..89\) (n số 8, n+1 số 4).
\(=9+8.\left(10+10^2+10^3+...+10^n\right)+4.\left(10^{n+1}+10^{n+2}+10^{n+3}+...+10^{2n+1}\right)\)
\(=9+8.10.\left(1+10+10^2+...+10^{n-1}\right)+4.10^{n+1}.\left(1+10+10^2+...+10^n\right)\)
\(=9+8.10.\dfrac{10^n-1}{9}+4.10^{n+1}.\dfrac{10^{n+1}-1}{9}\)
\(=9+\dfrac{8.10^{n+1}-80}{9}+\dfrac{4.10^{2n+2}-4.10^{n+1}}{9}\)
\(=\dfrac{81+8.10^{n+1}-80+4.10^{2n+2}-4.10^{n+1}}{9}\)
\(=\dfrac{4.10^{2n+2}-4.10^{n+1}+1}{9}=\left(\dfrac{2.10^{n+1}-1}{3}\right)^2\)
- Vì \(2.10^{n+1}\equiv2\left(mod3\right);1\equiv1\left(mod3\right)\)
\(\Rightarrow\left(2.10^{n+1}-1\right)⋮3\) nên \(\dfrac{2.10^{n+1}-1}{3}\) là số tự nhiên (vì \(n\in N\)).
- Vậy C là số chính phương.
