Question

Solve \(T(n)=T(n-1)+n\) for \(n \geq 1\) with \(T(0)=2\).

Short answer

The solution is \( T(n) = \frac{n(n+1)}{2} + 2 \).

Step-by-step solution

Identify the Type of Recurrence

The recurrence relation given is a linear recurrence relation with constant coefficients. It is defined as \( T(n) = T(n-1) + n \) with an initial condition \( T(0) = 2 \).

Consider Small Values

Compute the first few values manually to identify a pattern: \( T(0) = 2 \), \( T(1) = T(0) + 1 = 3 \), \( T(2) = T(1) + 2 = 5 \), \( T(3) = T(2) + 3 = 8 \), and so on. This might help recognize a pattern or solution form.

Formulate a General Hypothesis

Notice that the increasing sequence (2, 3, 5, 8, ...) looks similar to triangular numbers plus an offset. Hypothesize a general solution for \( T(n) = a_n + b \) where \( a_n = \frac{n(n+1)}{2} \) (triangular numbers). Analyze if any modification is needed.

Verify with Initial Condition

Checking \( T(n) = \frac{n(n+1)}{2} + 2 \): Verify by substitution into the recurrence relation. Check if \( T(0) = \frac{0(0+1)}{2} + 2 = 2 \), which satisfies the initial condition.

Substitute and Simplify to Confirm

Substitute the proposed formula \( T(n) = \frac{n(n+1)}{2} + 2 \) into the recurrence: For verification, \( T(n-1) + n = \left( \frac{(n-1)n}{2} + 2 \right) + n = \frac{n^2 - n + 2n}{2} + 2 = \frac{n(n+1)}{2} + 2 \). This holds true for all \( n \).

Conclusion

Thus, the general solution to the recurrence relation is \( T(n) = \frac{n(n+1)}{2} + 2 \).

Key concepts

Linear Recurrence Relation

A linear recurrence relation is a sequence of numbers in which each term is a linear function of its predecessors. In simple words, it's a way to calculate the next number in a sequence based on the previous ones using a specific formula.
Linear recurrence relations are important because they help predict future events based on a series of observations or conditions, making them really useful in various fields like computer science, finance, and more.
The recurrence relation given in the exercise is:

• Linear, which means it involves a sum and product with coefficients.
• Of order 1, since it depends only on the immediately prior term, namely, the term right before it, i.e., \( T(n) = T(n-1) + n \).

Recognizing the type of recurrence can help us find a solution efficiently, as different methods suit different types of recurrence relations.

Initial Condition

Initial conditions are the starting values of a sequence determined at the beginning of solving recurrence relations. They set the stage for calculating subsequent terms in the sequence.
For linear recurrence relations like the one in the exercise, if you don't know where to start, you can't determine other values. In simpler terms, the initial condition acts like the foundation of a house; it supports everything built on top of it.
In our case, the initial condition is \( T(0) = 2 \). This means when \( n = 0 \), the value of \( T \) is 2.

• This starting point allows us to compute further terms of the sequence.
• Using this initial value, we can begin finding patterns and formulating hypotheses for the sequence's behavior.

Ensuring the initial condition satisfies the derived sequence formula is critical as it confirms the hypothesized model is correct.

Triangular Numbers

Triangular numbers form a pattern of dot representations that create an equilateral triangle. Numerically, they sum up natural numbers in sequence. For instance, \( 1+2+3 = 6 \), forming a triangle-shaped figure with six dots.
The formula for the nth triangular number is \( \frac{n(n+1)}{2} \). Recognizing these numbers can reveal interesting patterns and solutions, especially in problems involving sequences.
In the given step-by-step solution:

• We notice the solution involves adding a specific offset to the formula for triangular numbers.
• The hypothesis guessed that the terms of the recurrence followed the shape of triangular numbers with an added constant.
• Thus, the solution takes the form \( T(n) = \frac{n(n+1)}{2} + 2 \).

Understanding this helps identify patterns and shortcuts when computing terms manually.

Verification of Solution

Verification is essential to ensure that the proposed solution consistently satisfies both the recurrence relation and the initial condition. It's like checking an answer to ensure accuracy.
To verify a solution, we substitute it back into the original recurrence relation and check if both sides match.
For our exercise:

• The proposed solution is \( T(n) = \frac{n(n+1)}{2} + 2 \).
• We validate it by substituting \( T(n-1) + n \) and confirming that it equates to \( \frac{n(n+1)}{2} + 2 \).
• We also check the initial condition to ensure \( T(0) = 2 \), which indeed holds true.

After verification, we confidently conclude the solution correctly models the given recurrence relation, providing reliable predictions for subsequent terms.