Question

Cho A= 3+2^2+2^3+2^4+...+2^98+2^99

      Tìm số tự nhiên n biết A+1=4^n

Answer 1

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

\(\Rightarrow A=1+2+2^2+...+2^{99}\)

\(\Rightarrow2A=2+2^2+...+2^{100}\)

\(\Rightarrow2A-A=2+2^2+...+2^{100}-1-2-...-2^{99}\)

\(\Rightarrow A=2^{100}-1\)

Thay A = 2¹⁰⁰ - 1 vào A + 1 = 4^n , ta có:

\(2^{100}-1+1=4^n\)

\(\Rightarrow2^{100}=2^{2n}\)

\(\Rightarrow2n=100\Rightarrow n=50\)