Question

[Phùng Ánh Dương]

Tìm số tự nhiên x biết : 

a, 1+3+5+7+9+...+(2x-1) = 225

b,\(2^x\)\(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2015}=2^{2019}-8\)^{\(2^{^{ }x}\)}

Answer 1

[Bui thi nhu quynh]

=> (1+2X-1)x (2x-1+1)/4=225

=> 2x+2x/4=225

=> 4x^2/4=225

=> x^2= 225

=> x=15

cái ^ là mũ nha bạn

chúc bn hok tốt

Answer 2

[Yen Nhi]

`Answer:`

a. Tổng: \([\left(2x-1\right)-1]:2+1=x\) số hạng

Ta có: \(1+3+5+7+9+...+\left(2x-1\right)=225\)

\(\Rightarrow x.\left(2x-1+1\right):2=225\)

\(\Leftrightarrow2x^2:2=225\)

\(\Leftrightarrow x^2=225\)

\(\Leftrightarrow x=15\)

b. Mình sửa đề nhé: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+...+2^{x+2015}=2^{2019}-8\)

\(\Rightarrow2^x.\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8\)

Ta đặt \(K=1+2+2^2+...+2^{2015}\)

\(\Rightarrow2^x.K=2^{2019}-8\)

\(\Rightarrow2K=2.\left(1+2+2^2+...+2^{2015}\right)\)

\(\Rightarrow2K=2+2^2+2^3+...+2^{2015}+2^{2016}\)

\(\Rightarrow2K-K=\left(2+2^2+2^3+...+2^{2015}+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)\)

\(\Rightarrow K=2^{2016}-1\)

\(\Rightarrow2^x.\left(2^{2016}-1\right)=2^{2019}-8\)

\(\Rightarrow2^{x+2016}-2^x=2^{2019}-2^3\)

\(\Rightarrow\hept{\begin{cases}x+2016=2019\\x=3\end{cases}}\Rightarrow x=3\)