Question

Solve \(T(n)=T(n-1)+5\) for \(n \geq 1\) with \(T(0)=1\).

Short answer

The solution is \( T(n) = 1 + 5n \) for \( n \geq 0 \).

Step-by-step solution

Understand the Recurrence Relation

The given recurrence relation is \( T(n) = T(n-1) + 5 \) with the initial condition \( T(0) = 1 \). This is a linear recurrence relation where each term is derived by adding 5 to the previous term.

Identify the Base Case

The base case is provided as \( T(0) = 1 \). This is the starting value of the sequence, which will be used to calculate subsequent terms.

Calculate First Few Terms

Using the base case, calculate the first few terms to identify a pattern: - \( T(1) = T(0) + 5 = 1 + 5 = 6 \)- \( T(2) = T(1) + 5 = 6 + 5 = 11 \)- \( T(3) = T(2) + 5 = 11 + 5 = 16 \)- Continue this pattern to see if a formula emerges.

Look for a Pattern or Formula

Notice the pattern:- \( T(1) = 1 + 1\times5 = 6 \)- \( T(2) = 1 + 2\times5 = 11 \)- \( T(3) = 1 + 3\times5 = 16 \)This can be generalized to: \( T(n) = 1 + n\times5 \).

Verify the Pattern

To ensure the pattern is correct, let's prove it by induction.- **Base Case**: For \( n=0 \), \( T(0) = 1 \) which matches \( 1 + 0\times5 = 1 \).- **Inductive Step**: Assume \( T(k) = 1 + k\times5 \). Show \( T(k+1) = 1 + (k+1)\times5 \). Starting from \( T(k+1) = T(k) + 5 = 1 + k\times5 + 5 = 1 + (k+1)\times5 \), the formula holds.

Key concepts

Base Case

In recurrence relations, the base case is an essential starting point. It provides the initial condition, which helps begin solving the entire sequence.
In our exercise, the base case is given as \( T(0) = 1 \). This specific value establishes the starting point for the sequence defined by our recurrence relation. Without it, we would have an infinitely many ways the sequence could begin.
The base case acts like the initial push of a domino, which leads to the domino effect of calculating each subsequent term.

Induction

Mathematical induction is a powerful tool for proving patterns and formulas in sequences and series. It is a step-by-step argument used in mathematical proofs.
In this exercise, we used induction to prove that the pattern \( T(n) = 1 + n \times 5 \) holds for all \( n \geq 0 \).

• Base Step: As already shown, \( T(0) = 1 \) squares with our formula \( 1 + 0 \times 5 = 1 \).
• Inductive Step: Assume the formula \( T(k) = 1 + k \times 5 \) is true for some arbitrary positive integer \( k \). We then need to show it holds for \( T(k+1) \). From earlier, since \( T(k+1) = T(k) + 5 = 1 + k \times 5 + 5 \), it is easy to see \( 1 + (k+1) \times 5 \), confirming its correctness.

This process of confirming via base and inductive steps ensures the formula we found isn't a fluke but a true representation of the pattern.

Linear Recurrence Relations

Linear recurrence relations are a type of recursive sequence where each term is a linear function of preceding terms.
For our given relation \( T(n) = T(n-1) + 5 \), each term is formed by adding a constant, 5, to the previous term \( T(n-1) \).
The constant nature of this addition tells us that the sequence is arithmetic, with equal differences (steps) between successive terms. This makes it easier to identify patterns and derive explicit formulas.
Linear recurrence relations rely heavily on previous terms and the base case to construct new terms and are a fundamental concept in algorithm analysis and problem-solving.