Question

Answer 1

Câu 1:
Đặt \(A=1+2+2^2+...+2^{2022}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2023}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2023}\right)-\left(1+2+2^2+...+2^{2022}\right)\)
\(\Rightarrow A=2^{2023}-1\)
\(\Rightarrow S=\dfrac{2^{2023}-1}{1-2^{2023}}=\dfrac{-\left(1-2^{2023}\right)}{1-2^{2023}}=-1\)

Answer 2

2:

a: B nguyên khi 2m+2+1 chia hết cho m+1

=>1 chia hết cho m+1

=>\(m+1\in\left\{1;-1\right\}\)

=>\(m\in\left\{0;-2\right\}\)

b: Gọi d=ƯCLN(2m+3;m+1)

=>2m+3-2m-2 chia hết cho d

=>1 chia hết cho d

=>d=1

=>ĐPCM