Bạn kiểm tra lại đề nhé, hình như đề hơi có vấn đề

\(\frac{3}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}\)

\(\Rightarrow\frac{3}{2}.A-A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}-\left(\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\right)\)

\(\Rightarrow\frac{1}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\Rightarrow A=2.\left(\frac{3}{2}\right)^{2013}-\frac{5}{2}\)

\(B-A=\frac{1}{2}.\left(\frac{3}{2}\right)^{2013}-2.\left(\frac{3}{2}\right)^{2013}+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)

Trần Thị Loan tại sao lại + 5/2?

ngu thế Bụng Mon

ủng hộ mình lên 280 điểm với các bạn

c) Cho B = (1.2.3....2012) . ( 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\) ) Chứng minh B chia hết cho 2013

B = (1.2.3....2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ...+ \(\dfrac{1}{2012}\) )

=(1.2.3...671...2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))

=(1.2.(3.671)...2012) . (1 + \(\dfrac{1}{2}\) +\(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))

=(1.2.2013...2012) . (1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2012}\))

Vậy B chia hết cho 2013

Đúng đấy, bạn cứ chép vào đi

Lời giải:
Ta có:

\(A-\frac{1}{2}=\frac{3}{2}+(\frac{3}{2})^2+...+(\frac{3}{2})^{2012}\)

\(\frac{3}{2}(A-\frac{1}{2})=(\frac{3}{2})^2+(\frac{3}{2})^3+....+(\frac{3}{2})^{2013}\\ \Rightarrow \frac{3}{2}(A-\frac{1}{2})-(A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}\)

$\Rightarrow \frac{1}{2}(A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}$

$A-\frac{1}{2}=2(\frac{3}{2})^{2013}-3$

$A=2(\frac{3}{2})^{2013}-2,5$

$\Rightarrow A-B=2(\frac{3}{2})^{2013}-2,5-(\frac{3}{2})^{2013}:2$

$=\frac{3}{2}(\frac{3}{2})^{2013}-2,5=(\frac{3}{2})^{2014}-2,5$