Question

[T h ả o_C H u B Y]

2² + 2² + 2³ + 2⁴ + ... + 2^x = 2²¹

^{Mn giúp mk nha!}

Answer 1

[Thiên Thần ( Fire Smoke Team )]

\(2^2+2^2+2^3+2^4+...+2^x=2^{21}\)

Đặt \(2^2+2^3+2^4+...+2^x\)là \(B\), ta có :

\(2B=2.\left(2^2+2^3+2^4+...+2^x\right)\)

\(\Rightarrow2B=2^3+2^4+2^5+...+2^{x+1}\)

\(\Rightarrow2B-B=\left(2^3+2^4+2^5+...+2^{x+1}\right)-\left(2^2-2^3+2^4+...+2^x\right)\)

\(\Rightarrow B=2^{x+1}-2^2\)

Thay \(B\)vào (1) ta có :

\(2^2+\left(2^{x+1}-2^2\right)=2^{21}\)

\(\Rightarrow2^{x+1}=2^{21}\)

\(\Rightarrow x+1=21\)

\(\Rightarrow x=21-1\)

\(\Rightarrow x=20\)

Vậy \(x=20\)