Question

Prove in \(E^{1}\) that (i) \(\sum_{k=1}^{n} k=\frac{1}{2} n(n+1) ;\) (ii) \(\sum_{k=1}^{n} k^{2}=\frac{1}{6} n(n+1)(2 n+1) ;\) (iii) \(\sum_{k=1}^{n} k^{3}=\frac{1}{4} n^{2}(n+1)^{2}\); (iv) \(\sum_{k=1}^{n} k^{4}=\frac{1}{30} n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)\).

Short answer

These identities are proved using mathematical induction for all four summation formulas.

Step-by-step solution

Prove (i) by Induction

To prove that \( \sum_{k=1}^{n} k = \frac{1}{2} n(n+1) \), we use mathematical induction. **Base Case**: For \( n = 1 \), the left side is \( 1 \) and the right side is \( \frac{1}{2} \times 1 \times (1+1) = 1 \). Thus, the base case holds. **Inductive Step**: Assume it holds for \( n = m \), i.e., \( \sum_{k=1}^{m} k = \frac{1}{2} m(m+1) \). Then, consider \( n = m + 1 \):\[ \sum_{k=1}^{m+1} k = \sum_{k=1}^{m} k + (m+1) = \frac{1}{2} m(m+1) + (m+1). \] Simplifying gives: \[ \frac{1}{2} m(m+1) + (m+1) = \frac{1}{2}(m+1)(m+2). \] Thus it holds for \( n = m + 1 \), completing the proof by induction.

Prove (ii) by Induction

Prove \( \sum_{k=1}^{n} k^{2} = \frac{1}{6} n(n+1)(2n+1) \) using induction. **Base Case**: For \( n=1 \), \( 1^2 = \frac{1}{6} \times 1 \times 2 \times 3 = 1 \). **Inductive Step**: Assume it's true for \( n=m \), \( \sum_{k=1}^{m} k^{2} = \frac{1}{6} m(m+1)(2m+1) \). For \( n = m+1 \), \[ \sum_{k=1}^{m+1} k^{2} = \sum_{k=1}^{m} k^{2} + (m+1)^2 \] = \( \frac{1}{6} m(m+1)(2m+1) + (m+1)^2 \). Simplifying gives \( \frac{1}{6}(m+1)(m+2)(2m+3) \). Hence, the formula is proven for \( n = m + 1 \) and by induction.

Prove (iii) by Induction

Prove \( \sum_{k=1}^{n} k^{3} = \frac{1}{4} n^{2}(n+1)^{2} \) using induction. **Base Case**: For \( n=1 \), \( 1^3 = \frac{1}{4} \times 1^2 \times 2^2 = 1 \). **Inductive Step**: Assume it's true for \( n=m \), \( \sum_{k=1}^{m} k^{3} = \frac{1}{4} m^{2}(m+1)^{2} \). For \( n = m+1 \), \( \sum_{k=1}^{m+1} k^{3} = \sum_{k=1}^{m} k^{3} + (m+1)^3 \), equals \( \frac{1}{4} m^{2}(m+1)^{2} + (m+1)^3 \). Simplifying gives \( \frac{1}{4} (m+1)^{2}(m+2)^{2} \). So, the formula holds for \( n = m + 1 \), completing the induction.

Prove (iv) by Induction

Prove \( \sum_{k=1}^{n} k^{4} = \frac{1}{30} n(n+1)(2n+1)(3n^2+3n-1) \) via induction. **Base Case**: For \( n=1 \), \( 1^4 = \frac{1}{30} \times 1 \times 2 \times 3 \times 5 = 1 \). **Inductive Step**: Assume it's true for \( n = m \), \( \sum_{k=1}^{m} k^{4} = \frac{1}{30} m(m+1)(2m+1)(3m^2+3m-1) \). For \( n = m+1 \), \( \sum_{k=1}^{m+1} k^{4} = \sum_{k=1}^{m} k^{4} + (m+1)^4 \). Operate the expansion, combining gives: \( \frac{1}{30} (m+1)(m+2)(2m+3)(3m^2+9m+5) \), or equivalently simplifying matches the provided form by induction. Thus, it holds for \( n = m+1 \).

Key concepts

Summation Formulas

Summation formulas are mathematical tools that allow us to calculate the total sum of a sequence of numbers following a specific pattern. They are incredibly useful, especially when dealing with large numbers of terms.
For example, the formula \( \sum_{k=1}^{n} k = \frac{1}{2} n(n+1) \) calculates the sum of the first \( n \) natural numbers.
These formulas take into account common patterns and relationships among numbers.

Similarly, for squared numbers, we use \( \sum_{k=1}^{n} k^2 = \frac{1}{6} n(n+1)(2n+1) \), and this helps us avoid tediously adding each term one by one.
These formulas make it easy to find the sum without manual addition.

Such formulas are derived through various methods including direct mathematical derivation or by using mathematical induction. Each has its utility, with induction being a powerful method to ensure correctness in a variety of scenarios.

Polynomial Expressions

Polynomial expressions are mathematical expressions that involve variables and coefficients, raised to whole number powers. These are common in many algebraic problems and can range from simple to complex.
For instance, consider the expression \( \frac{1}{2} n(n+1) \) used in the summation formula for the sum of natural numbers.
It is a polynomial of degree 2 as the highest power of \( n \) is squared.

Polynomial expressions can be linear, quadratic, cubic, or higher degree, depending on the problem.
In the exercise, various polynomial expressions like \( \frac{1}{6} n(n+1)(2n+1) \) and \( \frac{1}{4} n^2(n+1)^2 \) represent the sum of squares or cubes, respectively.
Learning to manipulate these expressions is key in algebra and helps solve both simple and complex mathematical problems efficiently.

Polynomial expressions form the building blocks of many mathematical concepts and formulas, making them a fundamental part of any math curriculum.

Base Case

The base case is the initial step in a mathematical induction proof. It establishes the truth of the statement for a specific starting value, often \( n = 1 \).
This step is crucial because it provides a firm foundation from which the inductive step can proceed.

For example, in the proof of \( \sum_{k=1}^{n} k = \frac{1}{2} n(n+1) \), the base case checks if the formula holds true when \( n = 1 \).
If the base case is true, it is like the first domino in a line, guaranteeing the subsequent proofs can follow logically.

Without an established base case, the entire reasoning could collapse since there would be no secure starting point.
It confirms that the formula works for the simplest situation and supports the logical argument for its truth in all scenarios addressed by induction.

Mastering the base case concept ensures accuracy and integrity in proofs whole-heartedly reliant on induction, a fundamental technique in mathematics.