To solve the recurrence relation given the initial conditions, use an iterative approach similar to the method used in Example 10.d) \( a_n = a_{n-1} + 2n + 3, a_0 = 4 \)(Insert iterative solution here) **Problem 31:** What is the value of each of these sums of terms of a geometric progression?### d) \(\sum_{j=0}^{8} 2 \cdot (-3)^j\)In this problem, we are asked to find the value of the sum of the first 9 terms (from \( j = 0 \) to \( j = 8 \)) of a geometric progression, where each term is given by \( 2 \cdot (-3)^j \).To find the value of this series, we can use the formula for the sum of the first \( n \) terms of a geometric progression:\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \]Where:- \( S_n \) is the sum of the first \( n \) terms.- \( a \) is the first term of the progression.- \( r \) is the common ratio.- \( n \) is the number of terms.For this series:- \( a = 2 \) (the first term when \( j = 0 \)).- \( r = -3 \) (the common ratio).- The series has 9 terms (\( n = 9 \)).Therefore, substituting these values into the formula:\[ S_9 = 2 \cdot \frac{1 - (-3)^9}{1 - (-3)} \]\[ S_9 = 2 \cdot \frac{1 - (-19683)}{1 + 3} \]\[ S_9 = 2 \cdot \frac{1 + 19683}{4} \]\[ S_9 = 2 \cdot \frac{19684}{4} \]\[ S_9 = 2 \cdot 4921 \]\[ S_9 = 9842 \]Thus, the value of the sum of the given geometric progression is **9842**.

Name: To solve the recurrence relation given the initial conditions, use an
Uploaded: 2024-03-05T11:53:49.000Z
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Description: To solve the recurrence relation given the initial conditions, use an iterative approach similar to the method used in Example 10.d) \( a_n = a_{n-1} + 2n + 3, a_0 = 4 \)(Insert iterative solution here) **Problem 31:** What is the value of each of t