Introduction

In the realm of algorithm analysis, recurrence relations often emerge as tools to describe the performance of recursive algorithms. The recurrence relation $T(n) = T(n - 1) + n$, in particular, is a quintessential example of an arithmetic sequence in algorithm analysis often seen when evaluating simple recursive algorithms. This article will detail the solution for this recurrence using technical explanations and examples.

Understanding the Recurrence

The recurrence $T(n) = T(n - 1) + n$ suggests that the value at $T(n)$ depends linearly on its predecessor $T(n - 1)$ with an additional increment of $n$. This is emblematic of an algorithm that adds a linearly increasing amount of work at each stage. Below is a step-by-step explanation to solve this recurrence with initial condition $T(1) = 1$:

Step-by-Step Solution

1. Setup of the Relation

Given the recurrence $T(n) = T(n - 1) + n$, we start with the base case $T(1) = 1$. The task is to express $T(n)$ explicitly without recursion.

2. Unrolling the Recurrence

• Start from the recurrence and unwind it:

$$\begin{align*} T(n) & = T(n - 1) + n \\ & = (T(n - 2) + (n - 1)) + n \\ & = ((T(n - 3) + (n - 2)) + (n - 1)) + n \\ & \vdots \\ & = 1 + 2 + \ldots + (n - 1) + n \end{align*}$$

• This clearly shows that $T(n)$ is equivalent to the sum of integers from $1$ to $n$.

3. Summation Formula

• The formula for the sum of the first $n$ natural numbers is:

$$T(n) = \frac{n(n + 1)}{2}$$

4. Verification

• Substitute $n = 1$ into the formula to verify the base condition:

$$T(1) = \frac{1(1 + 1)}{2} = 1$$

• Consider $n = 2$:

$$T(2) = \frac{2(2 + 1)}{2} = 3$$

• This matches the unrolled version as well: $$T(2) = 1 + 2 = 3$$

Evaluating Complex Scenarios

Given the simplicity, the solution enables quick calculations for larger $n$. For example, for $n = 10$, $T(10)$ would be:

$$T(10) = \frac{10 \times 11}{2} = 55$$

Technical Explanation

Time Complexity Implications

The expression $T(n) = \frac\{n(n + 1)\}\{2\}$ is $O(n^2)$. This quadratic complexity arises because each call incurs work proportional to $n$, leading to a cumulative work of the order $\frac{n^2}{2}$. As such, algorithms that follow this recurrence are not ideal for larger input sizes when optimum performance is critical.

Implementing in Code

Here is an example implementation of how this recurrence might be solved iteratively rather than recursively: