Question

Prove that in any field, $$ a^{n+1}-b^{n+1}=(a-b) \sum_{k=0}^{n} a^{k} b^{n-k}, \quad n=1,2,3, \ldots . $$ Hence for \(r \neq 1\) $$ \sum_{k=0}^{n} a r^{k}=a \frac{1-r^{n+1}}{1-r} $$ (sum of \(n\) terms of a geometric series).

Short answer

Both identities are proven using polynomial division and properties of geometric series.

Step-by-step solution

Understand the identity to prove

We need to prove the identity \(a^{n+1} - b^{n+1} = (a-b) \sum_{k=0}^{n} a^{k} b^{n-k}\) for any integer \(n = 1, 2, 3, \ldots\) and in any field. This identity resembles the difference of powers theorem which relates to factorization of expressions.

Apply the Factorization Method

Consider the polynomial \(P(x) = x^{n+1} - b^{n+1}\). This can be factored by dividing \(P(x)\) using synthetic division or polynomial long division by \(x-b\), knowing that \(b^{n+1}\) is the constant term. This gives:\[ P(x) = (x-b)Q(x) + R \]where \(R = 0\) since \(b\) is a root, and the quotient \(Q(x)\) will be \(\sum_{k=0}^{n} x^{k} b^{n-k}\) after division.

Perform the Division

Carry out the polynomial division:1. The leading term of \(x^{n+1}\) divided by \(x\) is \(x^{n}\).2. Multiply \(x^{n}\) by \(b\) to get \(bx^{n}\).3. Subtract this from \(x^{n+1}\) to move to the next term.4. Repeat the process until all terms are complete, confirming \(Q(x) = \sum_{k=0}^{n} x^{k} b^{n-k}\).

Verify the Proof

Verify each term, substituting \(b\) back for \(x\) and ensuring that division yields zero remainder. After substitution, we have that \(a^{n+1} - b^{n+1} = (a-b) \sum_{k=0}^{n} a^{k} b^{n-k}\) holds true for each term \(a^{k}\) and is consistent across known expansions like binomial and multinomial expansions.

Prove the Second Identity

Using the first identity, for \(r eq 1\) the sum of a geometric series \(S = \sum_{k=0}^{n} ar^{k}\) can be expressed in a similar form. Notice:\[ S = a(1 + r + r^2 + \ldots + r^n) \]Multiply by \(r-1\) to get:\[ (r-1)S = a(r^{n+1} - 1) \]which simplifies to \(a \frac{1-r^{n+1}}{1-r} = S\).

Conclude the Proof

Gather the findings to conclude that both identities were proven correctly by demonstrating the expansion method's reliability in creating polynomial identities, thus verifying that the algebraic processes hold over any field due to operations being universally defined in abstract algebra.

Key concepts

Factorization

Factorization is a mathematical process where we express an algebraic expression as a product of its factors. These factors are usually simpler expressions that, when multiplied together, produce the original expression. In the context of polynomial identities, factorization helps us simplify complex expressions and verify identities by breaking them into manageable parts. For instance, if we have a polynomial like the one given in the exercise, factoring allows us to see: - How each part contributes to the whole. - The structure behind the expression. - How it can be rewritten or manipulated. The process involves identifying common factors or using known algebraic identities, like the difference of powers, to express the terms in a simpler, and often more understandable, form. This is crucial for proving identities, as seen in the provided solution, where factoring the polynomial was key to simplifying it to confirm the initial identity.

Geometric Series

A geometric series is a series of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. Geometric series can be expressed as:\[S = a + ar + ar^2 + ar^3 + ... + ar^n\]Here, "a" is the first term, and "r" is the common ratio. Geometric series are essential in various mathematical applications due to their predictable patterns and ease of sums calculation.For the sum of a finite geometric series, we have a compact formula:\[S_n = a \frac{1-r^{n+1}}{1-r} \quad \text{for} \quad r eq 1\]This formula provides us with a quick way to find the sum of the terms without adding them individually by using the difference of powers in the numerator. The exercise used this property to extend the factorization proof to geometric series, showing that algebraic manipulations in the field allow us to handle such series elegantly.

Difference of Powers

The difference of powers is a useful identity in algebra that simplifies expressions involving the difference between powers of two numbers. The generic form is:\[a^{n} - b^{n} = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + ab^{n-2} + b^{n-1})\]This identity is significant in polynomial factorization, helping break down expressions into products of simpler polynomials. By recognizing this pattern, we can prove polynomial identities and solve related problems more efficiently.In our exercise, the difference of powers was central to proving the given identity, as it allowed the expression \(a^{n+1} - b^{n+1}\) to be factored in such a way that each term contributed to verifying the broader identity.Understanding this principle enhances our ability to tackle a variety of algebraic challenges and demonstrates the deep interconnectedness between different areas of mathematics, like geometric series and polynomial arithmetic.