Question

1.tính

a=1+2mu2+2mu4+2mu6.... +2mu96+2mu98+2mu100

Answer 1

Ta có; \(A=1+2^2+2^4+...+2^{100}\)

\(\Rightarrow4A=2^2A=2^2+2^4+...+2^{102}\)

\(\Rightarrow3A=4A-A=\left(2^2+...+2^{102}\right)-\left(1+...+2^{100}\right)\)

\(\Rightarrow3A=2^{102}-1\Rightarrow A=\frac{2^{102}-1}{3}\)