Tìm (x) biết (1 - 3 + 3^2 - 3^3 + ... + ( - 3)^x

Đặt \(A = 1 - 3 + {3^2} - {3^3} + ... + {\left( { - 3} \right)^x}\)

Khi đó \(3A = 3 - {3^2} + {3^3} - {3^4} + ... + {\left( { - 3} \right)^{x + 1}}\)

Suy ra \[A + 3A = \left[ {1 - 3 + {3^2} - {3^3} + ... + {{\left( { - 3} \right)}^x}} \right] + \left[ {3 - {3^2} + {3^3} - {3^4} + ... + {{\left( { - 3} \right)}^{x + 1}}} \right]\]

Do đó \(4A = 1 + {\left( { - 3} \right)^{x + 1}}\) nên \(A = \frac{{{{\left( { - 3} \right)}^{x + 1}} + 1}}{4}\)

Theo bài, \(A = 1 - 3 + {3^2} - {3^3} + ... + {\left( { - 3} \right)^x} = \frac{{{9^{1013}} + 1}}{4}\)

Suy ra \(\frac{{{{\left( { - 3} \right)}^{x + 1}} + 1}}{4} = \frac{{{9^{1013}} + 1}}{4}\)

Vì vậy \({\left( { - 3} \right)^{x + 1}} = {9^{1013}}\)

\({\left( { - 3} \right)^{x + 1}} = {3^{2026}}\)

\({\left( { - 3} \right)^{x + 1}} = {\left( { - 3} \right)^{2026}}\)

Suy ra \(x + 1 = 2026\), nên \(x = 2025\).

Vậy \(x = 2025\).