Question

Tinh        A= 3+33+333+......+33.....333

                                               { gom 50 so 3}  

Answer 1

\(3+33+333+...+33...333\)(50 chữ số 3)

\(=3.\frac{10^{50}+1-9.50-10}{81}\)

Answer 2

Theo đầu bài ta có:
\(A=3+33+333+...+33...333_{\left(50c.s\right)}\)
\(=9:3+99:3+999:3+...+99...999_{\left(50c.s\right)}:3\)
\(=\left\{\left(10-1\right)+\left(10^2-1\right)+\left(10^3-1\right)+...+\left(10^{50}-1\right)\right\}:3\)
\(=\left\{\left[10+10^2+10^3+...+10^{50}\right]-50\right\}:3\)
\(=\left\{\left[10\left(10+10^2+10^3+...+10^{50}\right)-\left(10+10^2+10^3+...+10^{50}\right)\right]:9-50\right\}:3\)
\(=\left\{\left[\left(10^2+10^3+10^4+...+10^{51}\right)-\left(10+10^2+10^3+...+10^{50}\right)\right]:9-450:9\right\}:3\)
\(=\left\{\left[\left(10\cdot10^{50}-10\cdot1\right)-10\cdot45\right]:9\right\}:3\)
\(=\left[10\cdot\left(10^{50}-1-45\right)\right]:27\)
\(=\frac{10}{27}\left(10^{50}-46\right)\)