Proving the Sum of Integers Formula Using Mathematical Induction

Mathematical induction is a powerful tool for proving statements or formulas in mathematics. In this article, we will explore how to prove the sum of the first n positive integers using induction. This technique can be extended to prove other formulas as well.

Introduction

The sum of the first n positive integers, denoted as ( S_n ), is given by the formula:

[ S_n 1 2 3 cdots n frac{n(n 1)}{2} ]

While this formula is well-known and easy to derive, let's use mathematical induction to prove its correctness, providing a deeper understanding of the method.

Proof by Induction

Mathematical induction involves three main steps:

Base Case: Verify the formula for a specific base value, typically ( n 0 ) or ( n 1 ). Inductive Hypothesis: Assume the formula is true for some arbitrary ( k ) in the interval [0, ( n )]. Inductive Step: Prove that if the formula is true for ( k ), then it must also be true for ( k 1 ).

Step 1: Base Case

Let's start with the base case, where ( n 0 ).

[text{For } n 0, quad S_0 0]

According to the formula:

[frac{0(0 1)}{2} 0]

Since both sides are equal, the formula holds for ( n 0 ).

Step 2: Inductive Hypothesis

Assume that the sum of the first ( k ) positive integers is given by:

[ S_k frac{k(k 1)}{2} ]

for some integer ( k geq 0 ).

Step 3: Inductive Step

We need to prove that if the formula is true for ( k ), then it must also be true for ( k 1 ).

Consider the sum of the first ( k 1 ) positive integers:

[text{For } n k 1, quad S_{k 1} 1 2 3 cdots k (k 1) ]

Using the inductive hypothesis, we can rewrite this as:

[text{For } n k 1, quad S_{k 1} S_k (k 1)] [text{Substitute } S_k text{ with the formula:}] [(k 1) frac{k(k 1)}{2} frac{2(k 1) k(k 1)}{2} frac{(k 1)(k 2)}{2} ]

Thus, if the formula is true for ( k ), it is also true for ( k 1 ).

Alternative Proof Using Telescopic Series

An alternative approach involves using a telescopic series to derive the sum. Consider the following identity:

[ 1 frac{(1 1k) - (k)}, quad text{for } k 0, 1, 2, ldots, n ]

Summing from ( k 0 ) to ( n ), we get:

[sum_{k0}^{n} left( frac{(1 1k) - k} right) n cdot 1 n cdot (0 1) ]

This expands to:

[sum_{k0}^{n} left( frac{(1 1k)}{2} - frac{k}{2} right) sum_{k0}^{n} left( frac{1}{2} frac{k}{2} - frac{k}{2} right) sum_{k0}^{n} 1 n 1 ]

Thus:

[sum_{k0}^{n} left( frac{1}{2} frac{k}{2} - frac{k}{2} right) sum_{k0}^{n} left( frac{1}{2} frac{k}{2} right) frac{n(n 1)}{2} ]

Conclusion

By using both mathematical induction and telescopic series, we have successfully proven the formula for the sum of the first ( n ) positive integers. This method can be extended to derive and prove other summation formulas for integer powers and other sequences.

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Keywords

mathematical induction, sum of integers, proof techniques