Chứng minh rằng A= 3/1×4+3/2×6+3/3×8+...+1/2012×1342 < 1,5
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
A=1+4+42+...+459A=1+4+42+...+459
A=(1+4+42)+(43+44+45)+...+(457+458+459)A=(1+4+42)+(43+44+45)+...+(457+458+459)
A=(1+4+42)+43(1+4+42)+...+447(1+4+42)A=(1+4+42)+43(1+4+42)+...+447(1+4+42)
A=21+43.21+...+447.21A=21+43.21+...+447.21
A=21(1+43+...+447)A=21(1+43+...+447)
⇒A⋮21
các số như 43,447,459,458........ là 4 mũ và các số đằng sau là số mũ
câu b cũng làm như vậy nhưng dổi các số và kết quả
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2012}{3^{2012}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{2012}{3^{2011}}\)
\(\Rightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{2012}{3^{2011}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2012}{3^{2012}}\right)\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2011}}-\frac{2012}{3^{2012}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2010}}-\frac{2012}{3^{2011}}\)
\(\Rightarrow6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2010}}-\frac{2012}{3^{2011}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2011}}-\frac{2012}{3^{2012}}\right)\)
\(\Rightarrow4A=3-\frac{2012}{3^{2011}}\)
\(\Rightarrow A=\frac{3-\frac{2012}{3^{2011}}}{4}=\frac{3}{4}-\frac{\frac{2012}{3^{2011}}}{4}=\frac{3}{4}-\frac{2012}{3^{2011}.4}\)
\(\Rightarrow A< \frac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(3+3^2+3^3+3^4\right)+3^4\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(=120+3^4.120+...+3^{2008}.120=120\left(1+3^4+...+3^{2008}\right)⋮120\)
\(A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(A=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(A=\left(3+3^2+3^3+3^4\right)\left(1+3^4+...+3^{2008}\right)\)
\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=3.40+3^5.40+...+3^{2009}.40\)
\(=120+3^4.120+...+3^{2008}.120\)
\(=120\left(1+3^4+...+3^{2008}\right)\)
Vì \(120⋮120\) nên \(120\left(1+3^4+...+3^{2008}\right)⋮120\)
hay \(A⋮120\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)