\(3A=100+\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}\)
\(3A-A=2A=100-\frac{100}{3^4}\)
\(A=50-\frac{\frac{100}{3^4}}{2}\)
\(\text{Đặt }A=\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}\)
\(3A=100+\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}\)
\(3A-A=2A=100-\frac{100}{3^4}\)
\(2A=100-\frac{100}{81}=\frac{8000}{81}\)
\(A=\frac{8000}{81}\text{ : }2\)
\(A=\frac{4000}{81}\)
Đặt \(A=\frac{100}{3}+\frac{100}{3^2}+\frac{100}{3^3}+\frac{100}{3^4}\)
\(\Rightarrow A=100\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\)
\(\Rightarrow3B-B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\)\(-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}\)
\(\Rightarrow2B=1-\frac{1}{3^4}\)
\(\Rightarrow B=\frac{1}{2}\left(1-\frac{1}{3^4}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{80}{81}=\frac{40}{81}\)
\(\Rightarrow A=100.\frac{80}{81}=\frac{8000}{81}\)
Ký hiệu \(\left[a\right]\) là phần nguyên của a có nghĩa là số nguyên lớn nhất ko vượt quá a.
\(VD:\left[3,5\right]=3;\left[3\right]=3;\left[-4,5\right]=-4\)
Quay trở lại bài toán,ta có:
\(A=33+11+3+1=48\)
