Let Sn represent the given statement, and use mathematical...

Let $S_{n}$ represent the given statement, and use mathematical induction to prove that $S_{n}$ is true for every positive integer $n$. (a) Verify $S_{1}$. (b) Write $S_{k}$. (c) Write $S_{k+1}$. (d) Assume that $S_{k}$ is true and use algebra to change $S_{k}$ to $S_{k+1}$. (e) Write a conclusion based on Steps $(a)-(d)$. $$3+9+27+\dots+3^{n}=\frac{1}{2}\left(3^{n+1}-3\right)$$

Let $S_{n}$ represent the given statement, and use mathematical induction to prove that $S_{n}$ is true for every positive integer $n$.
(a) Verify $S_{1}$.
(b) Write $S_{k}$.
(c) Write $S_{k+1}$.
(d) Assume that $S_{k}$ is true and use algebra to change $S_{k}$ to $S_{k+1}$.
(e) Write a conclusion based on Steps $(a)-(d)$.
$$3+9+27+\dots+3^{n}=\frac{1}{2}\left(3^{n+1}-3\right)$$

Key Concepts

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying equations using algebraic operations. In the context of induction proofs, such manipulation is often used to bridge the gap between the inductive hypothesis and the form required for demonstrating the statement for k+1.

Inductive Step

The inductive step involves using the inductive hypothesis (the assumption that the statement holds for k) to prove that the statement is valid for k+1. This step shows the logical progression from one case to the next, ensuring the property holds for all natural numbers.

Inductive Hypothesis

The inductive hypothesis assumes that the statement is true for some arbitrary positive integer k. This assumption is used as a stepping stone in demonstrating that the statement also holds for the next case, k+1, forming the core of the induction argument.

Base Case

The base case is the initial step in an inductive proof where the statement is verified to be true for the smallest value (usually n = 1). Establishing a valid base case is crucial because it serves as the foundation upon which the induction builds.

Mathematical Induction

Mathematical induction is a proof technique used to establish that a statement is true for all positive integers. It typically involves verifying the statement for a base case and then proving that if the statement holds for an arbitrary integer k, it must also hold for k+1, thus covering all possible cases.

Transcript

00:01 We are going to use mathematical induction to prove that the statement s -n is true for every positive integer n.

00:10 For that, we follow these steps.

00:13 In part a, we verify s -1.

00:16 That is proof that s -1 is true.

00:19 Part b, we write s -k.

00:21 In part c, we write s -k -plus -1.

00:25 In part -t, we assume that s -k is true and use algebra to change s -k to s -k -k -plus -1, and in part e, we write a conclusion -based.

00:33 On steps a through d.

00:36 The statement as n is the following.

00:40 3 plus 9 plus 27 plus up to 3 to the nth power is equal to 3 to the n plus 1 power, minus 3, all that divided by 2.

00:53 N is a natural number or a positive integer.

00:57 Before we start part a, we write the statement as n this way.

01:03 3 to the 1, that is 3 plus.

01:05 3 square which is 9 plus 3 to the 3 which is 27 and so on up to the term 3 to the nth power and that's equal to 3 to the m plus 1 power minus 3 all that over 2 and as we can see here the first term 3 to the 1 has exponent 1 and is the first term the second term has exponent 2 and it's the 2 term, the third term has exponent 3, and so on, up to the last term, the nth term, has exponent n.

01:50 So the position of the terms is determined by the exponent, so we are counting the terms using the exponent.

02:00 It means that we have in the sum, n terms.

02:04 So we can redefine or restate as n as the sum of the first n powers of 3 is equal to 3 to the n plus 1 minus 3 over 2.

02:23 So in part a, we're going to verify it's 1.

02:29 We have only one term.

02:30 It means we don't have a sum, but only the first term on the left side.

02:35 That is 3 to the 1...

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