91Ó°ÊÓ

In mathematical induction, the base case is the starting point of your proof. It is the first instance where we verify the formula or statement to be true. For our exercise, we need to check the formula for the smallest value of \(n\), which is \(n=1\).

When \(n=1\), the left-hand side of the equation is simply \(1\), as there is only one term, \(x^0 = 1\). The right-hand side becomes \(\frac{1-x^1}{1-x} = \frac{1-x}{1-x} = 1\). Since both sides are equal, the base case proves true.

A successful base case proves that our formula works for the very first number in our sequence of interest. Without a correct base case, induction cannot proceed, just like building a house on an unstable foundation.

The inductive hypothesis is the assumption that our statement or formula holds true for some arbitrary but fixed number, say \(n=k\).

In our problem, the inductive hypothesis assumes that this statement is valid for \(n=k\):

• \(1+x+x^2+\cdots+x^{k-1} = \frac{1-x^k}{1-x}\)

This assumption does not prove the statement for all numbers, but sets up the groundwork needed for the next file of the proof. By assuming it holds for \(n=k\), we look forward to verifying it also holds for \(n=k+1\). Without this hypothesis, it becomes difficult to show any progression beyond the base case.

In the inductive step, we use the inductive hypothesis to prove that if the statement holds for \(n=k\), then it must also hold for \(n=k+1\).

For this problem, consider the expression for \(n=k+1\):

• left side: \(1+x+x^2+\cdots+x^k\)

Using the inductive hypothesis, this sum becomes:\[1+x+x^2+\cdots+x^k = (1+x+x^2+\cdots+x^{k-1}) + x^k = \frac{1-x^k}{1-x} + x^k\]We manipulate this to show it equals the right side of the formula:\[\frac{1-x^k}{1-x} + x^k = \frac{1-x^k + x^k(1-x)}{1-x} = \frac{1-x^{k+1}}{1-x}\]The left side thus simplifies perfectly, completing the inductive step and moving the proof forward. This crucial step demonstrates the continuity and validity of the statement for all integers \(n\) beyond the base case.

The polynomial sum formula is a compact way to express the sum of a finite series of powers, particularly involving polynomials like the one we have been discussing.

For terms like \(1 + x + x^2 + \cdots + x^{n-1}\), the formula is expressed as:

• \(\frac{1-x^n}{1-x}\) where \(x eq 1\)

This formula simplifies calculating these sums significantly and is derived from the formula for the sum of a geometric series. Its elegance lies in the ability to handle large numbers of terms easily.

Understanding this formula helps in grasping the concept of series summation and is instrumental in exercises involving mathematical induction, as we've seen. It transforms an otherwise long and tedious addition into a single, manageable expression, showcasing the power of algebraic manipulation.