Questions: 1.3 Given: P = (1 - a) and T = (1 + a)(1 + a²)(1 + a⁴)...(1 + a⁵¹²) Determine the value of P × T in terms of a.

△ Determine the value of \( P \times T \) in terms of \( a \). ○ Express \( P \times T \) using the given definitions. ▷ Substitute \( P = (1-a) \) and \( T = (1+a)(1+a^2)(1+a^4)\dots(1+a^{512}) \) into \( P \times T \). ☼ \( P \times T = (1-a)(1+a)(1+a^2)(1+a^4)\dots(1+a^{512}) \). This combines \( P \) and \( T \) into a single product. ○ Apply the difference of squares formula iteratively. ▷ Use the formula \( (x-y)(x+y) = x^2 - y^2 \) repeatedly to simplify the product. ☼ First, \( (1-a)(1+a) = 1-a^2 \). Substituting, \( P \times T = (1-a^2)(1+a^2)(1+a^4)\dots(1+a^{512}) \).
Next, \( (1-a^2)(1+a^2) = 1-a^4 \). Substituting, \( P \times T = (1-a^4)(1+a^4)\dots(1+a^{512}) \).
Continuing this pattern:
\( (1-a^4)(1+a^4) = 1-a^8 \), so \( P \times T = (1-a^8)\dots(1+a^{512}) \).
\( (1-a^8)(1+a^8) = 1-a^{16} \), so \( P \times T = (1-a^{16})\dots(1+a^{512}) \).
\( (1-a^{16})(1+a^{16}) = 1-a^{32} \), so \( P \times T = (1-a^{32})\dots(1+a^{512}) \).
\( (1-a^{32})(1+a^{32}) = 1-a^{64} \), so \( P \times T = (1-a^{64})\dots(1+a^{512}) \).
\( (1-a^{64})(1+a^{64}) = 1-a^{128} \), so \( P \times T = (1-a^{128})\dots(1+a^{512}) \).
\( (1-a^{128})(1+a^{128}) = 1-a^{256} \), so \( P \times T = (1-a^{256})\dots(1+a^{512}) \).
Finally, \( (1-a^{256})(1+a^{256}) = 1-a^{512} \), and \( (1-a^{512})(1+a^{512}) = 1-a^{1024} \).
Thus, \( P \times T = 1-a^{1024} \). ✧ The value of \( P \times T \) is \( 1-a^{1024} \). ☺ $P \times T = 1 - a^{1024}$