Prove $3+9+27+...+3^n = \frac{3(3^n-1)}{2}$ By Induction

Hey math whizzes! Today, we're diving deep into the awesome world of mathematical induction. This technique is super handy for proving statements that hold true for all positive integers. We've got a specific statement, let's call it S(n), that we want to tackle:

$S(n): 3+9+27+\ldots+3^n = \frac{3(3^n-1)}{2}$

We're going to use the powerful tool of induction to show this statement is true for any positive integer $n$. Get ready, because we're breaking it down step-by-step!

Step 1: Base Case - Proving S(1)

Alright guys, the first thing we gotta do in induction is establish a base case. This means we check if our statement S(n) holds true for the smallest possible positive integer, which is $n=1$. So, let's plug in $n=1$ into our statement and see what happens.

On the left-hand side (LHS), we only have the first term, which is $3^1$. So, LHS = $3^1 = 3$.

Now, let's look at the right-hand side (RHS). We substitute $n=1$ into the formula:

RHS = $\frac{3(3^1-1)}{2}$

RHS = $\frac{3(3-1)}{2}$

RHS = $\frac{3(2)}{2}$

RHS = $\frac{6}{2}$

RHS = $3$

Boom! The LHS equals the RHS ($3 = 3$). This means our statement S(n) is true for $n=1$. We've successfully cleared the first hurdle! This base case is crucial because it gives us a starting point for our induction. Without a true base case, the entire inductive argument wouldn't hold water. So, pat yourselves on the back – we've got the foundation sorted!

Step 2: Inductive Hypothesis - Assume S(k) is True

Next up in our induction journey is the inductive hypothesis. This is where we make a pretty neat assumption: we assume that the statement S(n) is true for some arbitrary positive integer, let's call it $k$. So, for this integer $k$, we're going to pretend that the following equation is absolutely correct:

$S(k): 3+9+27+\ldots+3^k = \frac{3(3^k-1)}{2}$

Think of it like this: we're saying, "Okay, if this statement works for some number $k$, then we can use that fact to prove it works for the next number, $k+1$." This assumption is the engine that drives the inductive step. It's like giving ourselves a powerful tool, knowing it's true for a specific value, to help us build up the proof for all subsequent values. We don't need to prove S(k) here; we just need to assume it's true for the sake of moving forward. This assumption is the linchpin of the entire inductive proof, allowing us to bridge the gap from one integer to the next.

Step 3: Inductive Step - Prove S(k+1) is True

Now for the main event, the inductive step! This is where we use our inductive hypothesis (the assumption that S(k) is true) to prove that the statement must also be true for the next integer, $n = k+1$. Our goal is to show that S(k+1) holds:

$S(k+1): 3+9+27+\ldots+3^k+3^{k+1} = \frac{3(3^{k+1}-1)}{2}$

Let's start with the left-hand side (LHS) of S(k+1) and work our magic using algebra:

LHS of S(k+1) = $3+9+27+\ldots+3^k+3^{k+1}$

Look closely! The part $3+9+27+\ldots+3^k$ is exactly the LHS of our inductive hypothesis, S(k). And we assumed S(k) is true, which means we can replace that part with its RHS from the hypothesis:

LHS of S(k+1) = $\left(\frac{3(3^k-1)}{2}\right) + 3^{k+1}$

Now, we just need to use some good old-fashioned algebra to manipulate this expression and show it equals the RHS of S(k+1). Let's find a common denominator:

LHS of S(k+1) = $\frac{3(3^k-1)}{2} + \frac{2 \cdot 3^{k+1}}{2}$

Combine the numerators:

LHS of S(k+1) = $\frac{3 \cdot 3^k - 3 \cdot 1 + 2 \cdot 3^{k+1}}{2}$

LHS of S(k+1) = $\frac{3^{k+1} - 3 + 2 \cdot 3^{k+1}}{2}$

Now, notice that we have two terms with $3^{k+1}$. We can combine them: $3^{k+1} + 2 \cdot 3^{k+1} = 1 \cdot 3^{k+1} + 2 \cdot 3^{k+1} = (1+2) \cdot 3^{k+1} = 3 \cdot 3^{k+1}$

So, the expression becomes:

LHS of S(k+1) = $\frac{3 \cdot 3^{k+1} - 3}{2}$

Using the exponent rule $a^m \cdot a^n = a^{m+n}$, we can simplify $3 \cdot 3^{k+1}$ to $3^{k+1+1} = 3^{k+2}$. Wait, that's not quite right. Let's re-evaluate. We have $3 imes 3^{k+1}$, which is $3^1 imes 3^{k+1} = 3^{1+(k+1)} = 3^{k+2}$. Hmm, this doesn't seem to be leading to the target easily. Let's backtrack slightly on the simplification.

Let's go back to:

LHS of S(k+1) = $\frac{3^{k+1} - 3 + 2 \cdot 3^{k+1}}{2}$

We can factor out $3^{k+1}$ from the first and third terms in the numerator: $3^{k+1} + 2 \cdot 3^{k+1} = 3^{k+1}(1+2) = 3^{k+1}(3) = 3 imes 3^{k+1}$. This is correct. Now, let's rewrite the numerator using this:

LHS of S(k+1) = $\frac{3 imes 3^{k+1} - 3}{2}$

We can factor out a 3 from the numerator:

LHS of S(k+1) = $\frac{3(3^{k+1} - 1)}{2}$

And there you have it! This is exactly the RHS of our statement S(k+1). We started with the LHS of S(k+1) and, through algebraic manipulation and the use of our inductive hypothesis, we arrived at the RHS of S(k+1). This proves that if S(k) is true, then S(k+1) must also be true.

Step 4: Conclusion - S(n) is True for All Positive Integers n

We've made it to the finish line, guys! We've successfully completed all the steps of mathematical induction:

1. Base Case: We proved that the statement S(n) is true for $n=1$.
2. Inductive Hypothesis: We assumed that the statement S(n) is true for an arbitrary positive integer $k$.
3. Inductive Step: We used the inductive hypothesis to prove that the statement S(n) is also true for $n=k+1$.

Because we've shown that the statement is true for the initial case ($n=1$) and that it holds true for any subsequent case ($n=k+1$) if it's true for the preceding case ($n=k$), we can confidently conclude, by the principle of mathematical induction, that the statement $S(n)$ is true for all positive integers $n$.

So, the equation $3+9+27+\ldots+3^n = \frac{3(3^n-1)}{2}$ is officially proven for every positive integer $n$. Pretty cool, right? Induction is a seriously powerful tool in a mathematician's toolkit!