Question

[Nguyễn Như Quỳnh]

cho biểu thức B=\(\left(9^{2013}+9^{2012}+9^{2011}+...+9+1\right)\times200\)

CMR:B+25 là số chính phương

Answer 1

[Nghiêm Gia Phương]

Ta có: \(B=\left(9^{2013}+9^{2012}+9^{2011}+...+9+1\right)\times200\)

\(\Rightarrow9B=9\times\left(9^{2013}+9^{2012}+9^{2011}+...+9+1\right)\times200\)

\(\Rightarrow9B=\left(9^{2014}+9^{2013}+9^{2012}+...+9^2+9\right)\times200\)

\(\Rightarrow9B-B=\left(9^{2014}+9^{2013}+9^{2012}+...+9^2+9\right)\times200-\left(9^{2013}+9^{2012}+9^{2011}+...+9+1\right)\times200\)

\(\Rightarrow8B=\left\{\left(9^{2014}+9^{2013}+9^{2012}+...+9^2+9\right)-\left(9^{2013}+9^{2012}+9^{2011}+...+9+1\right)\right\}\times200\)

\(\Rightarrow8B=\left\{9^{2014}-1\right\}\times200\)

\(\Rightarrow8B=9^{2014}\times200-1\times200\)

\(\Rightarrow8B=9^{2014}\times200-200\)

\(\Rightarrow B=\frac{9^{2014}\times200-200}{8}\)

\(\Rightarrow B=\frac{9^{2014}\times200}{8}-\frac{200}{8}\)

\(\Rightarrow B=9^{2014}\times25-25\)

\(\Rightarrow B+25=9^{2014}\times25-25+25\)

\(\Rightarrow B+25=9^{2014}\times25\)

\(\Rightarrow B+25=9^{1007\times2}\times5^2\)

\(\Rightarrow B+25=\left(9^{1007}\right)^2\times5^2\)

\(\Rightarrow B+25=\left(9^{1007}\times5\right)^2\)

\(\Rightarrow B+25\) là số chính phương.

Vậy \(B+25\) là số chính phương (đpcm).

Answer 2

[Nguyễn Anh Duy]

Đặt \(A= 9^{2013}+9^{2012}+9^{2011}+...+9+1\)
\(\Longrightarrow 9A = 9^{2014} + 9^{2013} + 9^{2012} + 9^2 + 9\)
\(\Longrightarrow 8A = 9A - A = (9^{2014} + 9^{2013} + 9^{2012} + 9^2 + 9) - (9^{2013}+9^{2012}+9^{2011}+...+9+1) = 9^{2014} - 1\)
\(\Longrightarrow B= 200A = 25(9^{2014} - 1) = 25.9^{2014} - 25\)
\(\Longrightarrow B + 25 = 25.9^{2014} = (5.9^{1007})^2\)
\(\Longrightarrow B\) là số chính phương

Ngắn gọn nhé :)