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Đặt \(C=B-\frac{1}{2}=\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2012}\)
\(\Rightarrow\frac{3}{2}\cdot C=\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow\frac{3}{2}C-C=\frac{1}{2}C=\frac{3}{2}+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow C=3+\left(\frac{3}{2}\right)^{2013}\cdot2\)
\(\Rightarrow B=\frac{1}{2}+3+\left(\frac{3}{2}\right)^{2013}\cdot2\)
do đó \(A-B=\left(\frac{3}{2}\right)^{2014}+\frac{7}{2}\)
\(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2012}\)(1)
\(\frac{3}{2}A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2013}\)(2)
Lấy (2) trừ (1) ta được:
\(\frac{1}{2}A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\)
\(A=\frac{\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}}{\frac{1}{2}}=\left(\frac{3}{2}\right)^{2013}.2-\frac{5}{4}.2=\left(\frac{3}{2}\right)^{2013}.2-\frac{5}{2}\)
\(\Rightarrow B-A=\left(\frac{3}{2}\right)^{2013}\cdot\frac{1}{2}-\left(\frac{3}{2}\right)^{2013}.2+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)
Ta có \(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\)
\(\Rightarrow\frac{3}{2}A=\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+....\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow\frac{3}{2}A-A=\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\)hay \(\frac{1}{2}A=\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\)
Suy ra \(A=2.\text{[}\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\text{]}\)
Khi đó \(B-A=\frac{\left(\frac{3}{2}\right)^{2013}}{2}-2.\text{[}\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}\text{]}\)
\(A=\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\)
\(\frac{3}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}\)
\(\Rightarrow\frac{3}{2}.A-A=\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+...+\left(\frac{3}{2}\right)^{2013}-\left[\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+...+\left(\frac{3}{2}\right)^{2012}\right]\)
\(\Rightarrow\frac{1}{2}.A=\frac{3}{4}+\left(\frac{3}{2}\right)^{2013}-\frac{1}{2}-\frac{3}{2}=\left(\frac{3}{2}\right)^{2013}-\frac{5}{4}\)
\(\Rightarrow A=2.\left(\frac{3}{2}\right)^{2013}-\frac{5}{2}\)
\(B-A=\frac{1}{2}.\left(\frac{3}{2}\right)^{2013}-2.\left(\frac{3}{2}\right)^{2013}+\frac{5}{2}=-\left(\frac{3}{2}\right)^{2014}+\frac{5}{2}\)
Từng bài 1 thôi nha bn!!!
a) Xét hiệu: A = 9.(7x+4y) - 2. (13x+18y)
A = 63x + 36y - 26x - 36y
A = 37x \(\Rightarrow A⋮37\) Vì 7x + 4y chia hết cho 37
9.(7x+4y) chia hết cho 37
Mà A chia hết cho 37
\(2\left(13x+18y\right)⋮37\)
Do 2 và 37 là nguyên tố cùng nhau
13x+18y chia hết cho 37
Vậy nếu 7x+4y chia hết cho 37 thì 13x+18y chia hết cho 37
theo công thức \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
=>\(A=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+\frac{1}{4}.\frac{4.5}{2}+...+\frac{1}{2013}.\frac{2013.2014}{2}\)
\(=>A=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2014}{2}=>A=\frac{1}{2}\left(1+2+3+..+2014\right)-\frac{1}{2}\)
\(=>A=\frac{1}{2}.\frac{2014.2015}{2}-\frac{1}{2}=1014552\)
