Given P = (1-a) and T = (1+a)(1+a2)(1+a4)...(1+a512). Determine the value of P * T. Also, consider the geometric series 4 + 2 + 1 + 1/2 + ... Does it converge? Justify.

Alright ~👻Masana — let's do this. 1.3 Given P = (1-a) and T = (1+a)(1+a^2)(1+a^4)...(1+a^512). We need to determine the value of P × T. Step 1: Write out the product P × T. P × T = (1-a)(1+a)(1+a^2)(1+a^4)...(1+a^512) Step 2: Apply the difference of squares formula, (x-y)(x+y) = x^2 - y^2, repeatedly. The first two terms (1-a)(1+a) simplify to (1-a^2). P × T = (1-a^2)(1+a^2)(1+a^4)...(1+a^512) Next, (1-a^2)(1+a^2) simplifies to (1-(a^2)^2) = (1-a^4). P × T = (1-a^4)(1+a^4)...(1+a^512) This pattern continues. The exponents of a in the factors of T are 2^0, 2^1, 2^2, ..., 2^9. The last term in the product is (1+a^512), where 512 = 2^9. The product will continue until the last term is multiplied. The term before (1+a^512) would have been (1-a^256). So, the final step in the pattern is (1-a^512)(1+a^512). (1-a^512)(1+a^512) = 1-(a^512)^2 = 1-a^1024 Step 3: State the final simplified expression. P × T = 1-a^1024 2.1.1 Consider the geometric series: 4 + 2 + 1 + (1)/(2) + ... Does this series converge? Justify your answer. Step 1: Identify the first term (a) and the common ratio (r). The first term is a = 4. The common ratio is r = (2)/(4) = (1)/(2). Step 2: Check the condition for convergence of a geometric series. A geometric series converges if and only if the absolute value of its common ratio is less than 1, i.e., |r| < 1. In this case, |r| = |(1)/(2)| = (1)/(2). Step 3: Justify the convergence. Since (1)/(2) < 1, the series converges. 2.1.2 Calculate S_. Step 1: Use the formula for the sum to infinity of a convergent geometric series. The formula is S_ = (a)/(1-r). From 2.1.1, we have a=4 and r=(1)/(2). Step 2: Substitute the values and calculate S_. S_ = (4)/(1 - 1)2 S_ = (4)/(1)2 S_ = 4 × 2 S_ = 8 2.2 Given: _p=k^10 3^p-1 = 29520. Calculate the value of k. Step 1: Identify the properties of the geometric series. The sum is _p=k^10 3^p-1. The first term of this sum is when p=k, so A = 3^k-1. The common ratio is R = 3. The number of terms is N = 10 - k + 1. Step 2: Write the sum using the formula for a geometric series. The sum of a geometric series is S_N = (A(R^N - 1))/(R-1). 29520 = 3^k-1(3^(10-k+1) - 1)3-1 29520 = 3^k-1(3^11-k - 1)2 Step 3: Simplify the equation. Multiply both sides by 2: 59040 = 3^k-1(3^11-k - 1) Distribute 3^k-1: 59040 = 3^k-1 · 3^11-k - 3^k-1 Using the exponent rule x^a · x^b = x^a+b: 59040 = 3^(k-1) + (11-k) - 3^k-1 59040 = 3^10 - 3^k-1 Step 4: Calculate 3^10 and solve for 3^k-1. 3^10 = 59049 Substitute this value into the equation: 59040 = 59049 - 3^k-1 Rearrange to solve for 3^k-1: 3^k-1 = 59049 - 59040 3^k-1 = 9 Step 5: Express 9 as a power of 3 and solve for k. 3^k-1 = 3^2 Equating the exponents: k-1 = 2 k = 2 + 1 k = 3 Send me the next one 📸