d) an — — 4 аn-1 + 2n + 3, ао

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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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)
Transcribed Image Text: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**.
Transcribed Image Text:**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**.
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