91Ó°ÊÓ

In mathematical induction, the base case is the starting point. It is where we first check if the given statement is true for the smallest natural number, often \( n = 1 \). This first step is crucial because it sets the foundation for the rest of the proof. If the base case holds, it suggests a possibility that the statement may be true for all subsequent numbers.
For example, in the given exercise, we substitute \( n = 1 \) into the formula: \(1^2 = \frac{1(1+1)(2\times1 + 1)}{6} \). After simplifying both sides of the equation, they equal \( 1 \). Since both sides are equal, the base case is shown to be true. This confirmation is essential before moving to the next stages of induction.

The inductive hypothesis is a critical step where we assume the statement is true for some arbitrary natural number \( k \). It is similar to the premise where we take a leap of faith in the assumption and use this to extend the truth to the next number.
In our specific problem, assuming the hypothesis means believing that this equation holds: \( 1^2 + 2^2 + \, ... \, + k^2 = \frac{k(k+1)(2k+1)}{6} \). This assumed truth is the core of the induction process, serving as a bridge to demonstrate the truth for the next integer, \( k+1 \).
It is essential to clearly state and understand the inductive hypothesis because future steps depend on its assumed validity.

The inductive step is the heart of mathematical induction. It's where we prove the statement holds for \( k+1 \), building on the assumption that it holds for \( k \) (our inductive hypothesis). This step involves mathematical transformation.
In our example, to prove: \( 1^2 + 2^2 + \, ... \, + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} \), we start from \( 1^2 + 2^2 + \, ... \, + k^2 + (k+1)^2 \), and use the inductive hypothesis to replace part of the equation: \( \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \).
Through a series of mathematical manipulations, we aim to reformulate this to match the expression for \( k+1 \) in the given statement. This linkage from \( k \) to \( k+1 \) is what extends the original truth to cover all integers.

Simplification is an integral part of mathematical proofs, especially in induction. It involves breaking down and reassembling expressions to illustrate their equivalency. Simplification makes the proof persuasive and transparent.
In proving the given statement, after using the inductive hypothesis, we obtained an expression that needed manipulation: \( \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} \). Through arithmetic operations, including adding fractions and factoring, our goal was to adjust the expression to resemble the statement for \( k+1 \): \( \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} \).
Simplifying complex expressions into recognizable ones helps confirm that what works for \( k \) also does so for \( k+1 \), thus completing the inductive proof in a clear and concise manner.