Google
Câu hỏi: Tìm số tự nhiên n thỏa mãn
\(C_{2n}^0 + C_{2n}^2 + C_{2n}^4... + C_{2n}^{2n} = {2^{2021}}\)
\(C_{2n}^0 + C_{2n}^2 + C_{2n}^4... + C_{2n}^{2n} = {2^{2021}}\)
Lời giải chi tiết
\({(1 + x)^{2n}} = C_{2n}^0 + C_{2n}^1x + C_{2n}^2{x^2} + ... + C_{2n}^{2n}{x^{2n}}\)
Thay \(x = 1\) vào hai vế, ta suy ra
\(C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n} = {2^{2n}}\)
Thay \(x = - 1\) vào hai vế, ta suy ra
\(C_{2n}^0 - C_{2n}^1 + C_{2n}^2 - ... + C_{2n}^{2n} = 0\)
\(\begin{array}{l} \Rightarrow \left( {C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n}} \right) + \left( {C_{2n}^0 - C_{2n}^1 + C_{2n}^2 - ... + C_{2n}^{2n}} \right) = {2^{2n}}\\ \Leftrightarrow 2\left( {C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n}} \right) = {2^{2n}}\\ \Leftrightarrow C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n} = {2^{2n - 1}}\\ \Leftrightarrow 2n - 1 = 2021\\ \Leftrightarrow n = 1011\end{array}\)
\({(1 + x)^{2n}} = C_{2n}^0 + C_{2n}^1x + C_{2n}^2{x^2} + ... + C_{2n}^{2n}{x^{2n}}\)
Thay \(x = 1\) vào hai vế, ta suy ra
\(C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n} = {2^{2n}}\)
Thay \(x = - 1\) vào hai vế, ta suy ra
\(C_{2n}^0 - C_{2n}^1 + C_{2n}^2 - ... + C_{2n}^{2n} = 0\)
\(\begin{array}{l} \Rightarrow \left( {C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n}} \right) + \left( {C_{2n}^0 - C_{2n}^1 + C_{2n}^2 - ... + C_{2n}^{2n}} \right) = {2^{2n}}\\ \Leftrightarrow 2\left( {C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n}} \right) = {2^{2n}}\\ \Leftrightarrow C_{2n}^0 + C_{2n}^1 + C_{2n}^2 + ... + C_{2n}^{2n} = {2^{2n - 1}}\\ \Leftrightarrow 2n - 1 = 2021\\ \Leftrightarrow n = 1011\end{array}\)