Đặt B=9+999+99999+...+999...9(99 chữ số 9)
\(=10-1+10^3-1+10^5-1+\cdots+10^{99}-1\)
\(=\left(10+10^3+\cdots+10^{99}\right)-50\)
Đặt \(C=10+10^3+\cdots+10^{99}\)
=>\(100C=10^3+10^5+\cdots+10^{101}\)
=>\(100C-C=10^3+10^5+\cdots+10^{101}-10-10^3-\cdots-10^{99}\)
=>\(99C=10^{101}-10\)
=>\(C=\frac{10^{101}-10}{99}\)
Ta có: \(B=\left(10+10^3+\cdots+10^{99}\right)-50\)
\(=\frac{10^{101}-10}{99}-50=\frac{10^{101}-10-50\cdot99}{99}=\frac{10^{101}-4960}{99}\)
Ta có: A=3+333+...+33...3(99 chữ số 3)
\(=\frac13\left(9+999+99999+\cdots+999\ldots9\right)\) (99 chữ số 9)
\(=\frac13\cdot B=\frac13\cdot\frac{10^{101}-4960}{99}=\frac{10^{101}-4960}{297}\)
