91Ó°ÊÓ

In mathematical induction, the base case is the foundation of your proof. It is where you verify the given statement for the smallest possible value of the variable, usually starting with n=1. This step ensures that the statement holds true at the beginning of the sequence.
To demonstrate this concept, let's consider the given statement: $$3+9+27+\text{...}+3^{n}=\frac{1}{2}(3^{n+1}-3)$$.
For the base case, we set n=1:
- The left side of the equation becomes 3.
- The right side of the equation evaluates to \frac{1}{2}(3^{1+1} - 3) = \frac{1}{2} \times (9 - 3) = \frac{1}{2} \times 6 = 3.
Since the left side equals the right side, the statement holds true for n=1.
This is critical because proving the base case correctly is the first step that allows us to build the entire proof using induction.

The inductive step is the heart of mathematical induction. It involves two parts: the induction hypothesis and the induction argument.
First, you assume the statement holds for some arbitrary positive integer k. This assumption is called the induction hypothesis. For our example:
$$3+9+27+\text{...}+3^{k}=\frac{1}{2}(3^{k+1} - 3)$$.
This assumption allows us to proceed to the next part of the inductive step, which is to show that if the statement holds for n=k, then it also holds for n=k+1.
We need to prove:
$$3+9+27+\text{...}+3^k + 3^{k+1} = \frac{1}{2}(3^{k+2} - 3).$$
Using the induction hypothesis, we start with:
$$3 + 9 + 27 + \text{...} + 3^k = \frac{1}{2}(3^{k+1} - 3)$$
By adding \(3^{k+1}\) to both sides:
$$3 + 9 + 27 + \text{...} + 3^k + 3^{k+1} = \frac{1}{2}(3^{k+1} - 3) + 3^{k+1}.$$
This shows the structure for how we can move from n=k to n=k+1, crucially weaving the induction hypothesis into the proof and verifying that the statement holds for all values following the base case.

Algebraic manipulation is essential in the inductive step, enabling you to transform and simplify expressions to prove the statement for n=k+1.
Continuing from our induction assumption, after adding \(3^{k+1}\) to both sides: $$3+9+27+\text{...}+3^k + 3^{k+1} = \frac{1}{2}(3^{k+1} - 3) + 3^{k+1}.$$
To simplify the right side, factor out \frac{1}{2} and combine terms:
$$\frac{1}{2}(3^{k+1} - 3) + \frac{2}{2}(3^{k+1}) = \frac{1}{2}(3^{k+1} - 3 + 2 \times 3^{k+1}).$$
Then,
$$\frac{1}{2}(3^{k+1} - 3 + 2 \times 3^{k+1}) = \frac{1}{2}(3 \times 3^{k+1} - 3) = \frac{1}{2}(3^{k+2} - 3).$$
Notice how we simplified the equation by combining like terms and factoring out common elements.
These manipulations ensure the newly obtained expression for \(n=k+1\) holds, given the induction hypothesis.
Correct use of algebraic manipulation thus solidifies the inductive proof. By confirming the expression transformation through algebra, you effectively demonstrate that the original statement runs true for every subsequent integer in the sequence.