Math Problem Statement
Find the general term of the sequence based on the given recurrence relation.
Solution
This problem is asking to find the general term of the sequence based on the recurrence relation provided:
b_0 = 4, \\ b_n = b_{n-1} + 2n + 4, \quad \forall n \in \mathbb{N}^*. \end{cases}$$ Instead of using backward substitution or induction, we can apply a formula to deduce the general term. Here’s the solution: --- ### Step 1: Structure of the recurrence The recurrence equation can be expanded to analyze the pattern. Write a few terms: - $$b_0 = 4$$, - $$b_1 = b_0 + (2 \cdot 1 + 4) = 4 + 6 = 10$$, - $$b_2 = b_1 + (2 \cdot 2 + 4) = 10 + 8 = 18$$, - $$b_3 = b_2 + (2 \cdot 3 + 4) = 18 + 10 = 28$$. --- ### Step 2: Generalize using summation The recurrence relation implies: $$b_n = b_0 + \sum_{k=1}^n (2k + 4).$$ Substituting $$b_0 = 4$$: $$b_n = 4 + \sum_{k=1}^n (2k + 4).$$ --- ### Step 3: Separate the summation The summation $$\sum_{k=1}^n (2k + 4)$$ can be split into: $$\sum_{k=1}^n (2k + 4) = \sum_{k=1}^n 2k + \sum_{k=1}^n 4.$$ 1. For $$\sum_{k=1}^n 2k$$: $$\sum_{k=1}^n 2k = 2 \sum_{k=1}^n k = 2 \cdot \frac{n(n+1)}{2} = n(n+1).$$ 2. For $$\sum_{k=1}^n 4$$: $$\sum_{k=1}^n 4 = 4n.$$ --- ### Step 4: Simplify the summation Combine the results: $$\sum_{k=1}^n (2k + 4) = n(n+1) + 4n.$$ Simplify further: $$\sum_{k=1}^n (2k + 4) = n^2 + 5n.$$ Thus: $$b_n = 4 + n^2 + 5n.$$ --- ### Final Answer: $$b_n = n^2 + 5n + 4.$$ Let me know if you would like additional clarification or details! 😊 --- ### 5 Related Questions: 1. What is the value of $$b_5$$ in this sequence? 2. How does summation notation help in solving recurrence relations? 3. Can you verify $$b_n = n^2 + 5n + 4$$ using induction? 4. What would the general term be if the recurrence relation was $$b_n = b_{n-1} + 3n + 2$$? 5. How do recurrence relations relate to arithmetic and geometric sequences? ### Tip: For sequences defined recursively, always analyze the sum of increments to identify patterns in the general term!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Recurrence Relations
Summation
Sequences
Algebra
Formulas
b_n = b_0 + \sum_{k=1}^n (2k + 4)
Sum of first n integers: \sum_{k=1}^n k = \frac{n(n+1)}{2}
Sum of constant: \sum_{k=1}^n c = cn
Theorems
Recurrence Relation Expansion
Summation of Arithmetic Sequences
Suitable Grade Level
Grades 10-12