Question

So Sánh

3+3²+...+3⁹⁹  và 3¹⁰⁰

 

Answer 1

\(3+3^2+.......+3^{99}\)

\(\Rightarrow3\left(3+3^2+.....+3^{99}\right)=3^2+3^3+......+3^{100}\)

\(\Rightarrow\left(3^2+3^3+.....+3^{100}\right)-\left(3+3^2+....+3^{99}\right)\)

\(=3^{100}-3\)

\(Do3^{100}-3< 3^{100}=>3+3^2+....+3^{99}< 3^{100}\)

Answer 2

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 \(B=3^{100}\)

\(\Rightarrow A< B\)

Answer 3

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