\(3+3^2+.......+3^{99}\)
\(\Rightarrow3\left(3+3^2+.....+3^{99}\right)=3^2+3^3+......+3^{100}\)
\(\Rightarrow\left(3^2+3^3+.....+3^{100}\right)-\left(3+3^2+....+3^{99}\right)\)
\(=3^{100}-3\)
\(Do3^{100}-3< 3^{100}=>3+3^2+....+3^{99}< 3^{100}\)
Đặt
\(A=3+3^2+...+3^{99}\)
\(B=3^{100}\)
\(3A=3^2+3^3+...+3^{100}\)
\(3A-A=\left(3^2+3^3+...+3^{100}\right)-\left(3+3^2+...+3^{99}\right)\)
\(2A=3^{100}-3\)
\(A=\frac{3^{100}-3}{2}\) mà \(B=3^{100}\)
\(\Rightarrow A< B\)
cho tổng của 3+3^2+3^3+....+3^99 là C
Ta có: C= 3+3^2+3^3+.....+3^99 (1)
3C=3^2+3^3+3^4+...+3^100 (2)
Từ (1); (2): ta tối giãn 3^2;3^3;3^4;......;3^99
3C-C= 3^100-3
=> 3+3^2+3^3+3^4+....+3^99 < 3^100
