Problem 61

Question

True or False? In Exercises 61 and 62, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The sum of the first $n$ positive integers is $n(n+1) / 2$.

Step-by-Step Solution

Verified

Answer

The statement is true.

1Step 1: Understand the formula

The formula $n(n+1) / 2$ is a way to find the sum of the first $n$ positive integers. For example, if $n = 4$, then the sum of the first 4 positive integers would be $1 + 2 + 3 + 4 = 10$, and if we apply the formula it would be $4(4+1) / 2 = 10$. So it seems to work in this case.

2Step 2: Prove the formula

To definitively determine whether the formula is true or false, we need to prove it. We can use the principle of mathematical induction to do this. We first verify that the expression holds for $n = 1$. The left side (sum of the first 1st positive integers) is 1, and the right side (applying the formula) is also 1. Then, we assume that the formula holds for some positive integer $k$, and aim to show that it holds for $k + 1$, i.e. $1 + 2 + ... + k + (k + 1) = (k(k + 1)/2) + (k + 1)$. After simplifying, we end up with $(k+1)((k+2)/2)$, which is indeed the right side of the equation for $n = k + 1$, thus the formula is true. This fulfills the requirements for an inductive proof, hence the statement is true.

Key Concepts

Sum of Integers FormulaProof by InductionAlgebraic Manipulation

Sum of Integers Formula

When you see the formula $ \frac{n(n+1)}{2} $, you're looking at a handy tool to calculate the sum of the first $n$ positive integers. Why does this formula work, and how can it simplify your mathematical life? Let's break it down.

Consider the series $1, 2, 3, \ldots, n$. This formula sums them up easily without manually adding each term.
The formula $\frac{n(n+1)}{2}$ applies the pattern of summation and algebraic simplification. It efficiently condenses sums into a simple expression.
When $n$ is small, you might not need a formula, but imagine calculating the sum up to 100! The formula simplifies drastically.

To derive why $\frac{n(n+1)}{2}$ works, think about pairing numbers: the first and last term (1 and $n$), the second and second-to-last term (2 and $n-1$), and so on. Each pair sums to $n + 1$. If $n$ is even, there are exactly $\frac{n}{2}$ such pairs. If odd, there's an extra middle term, but overall, once simplified, you get an elegant $\frac{n(n+1)}{2}$. This approach makes finding the sum not only faster but also clearer.

Proof by Induction

Mathematical Induction is like climbing a ladder: once you secure the first step (base case), proving the rest is all about showing the method holds as you step higher. Let's look at how we do this with our sum of integers formula.

Base case: Start with $n=1$. The sum of the first integer is obviously 1, and the formula $\frac{1(1+1)}{2}$ also gives 1. Great, it works for $n=1$!
Inductive step: Assume the formula holds for a generic number $k$. Suppose it works for $1 + 2 + \ldots + k = \frac{k(k+1)}{2}$. Then ensure it holds for $k+1$.
For $k+1$, look at $1 + 2 + \ldots + k + (k+1)$. By the assumption, this is $\frac{k(k+1)}{2} + (k+1)$.

Algebraic manipulation tips in our favor: rewrite and simplify the expression. Expand your brackets and combine terms carefully. You get $\frac{(k+1)(k+2)}{2}$, which is exactly the formula adapted for $n=k+1$. Therefore, our induction is complete, and the formula is proven true for all positive integers.

Algebraic Manipulation

Algebraic manipulation is your skillful toolkit in mathematics, helping to transform and simplify equations. Let's delve into how it supports mathematical induction and the proof of formulas.

Algebraic Manipulation enhances clarity. When you manipulate expressions, you're often transforming them into more usable forms. This involves expanding, factoring, or collecting like terms.
Consider the inductive step in our proof: $\frac{k(k+1)}{2} + (k+1)$. It's crucial to streamline this. The key lies in having all terms share a common structure.
Factor out common terms; in this case, notice how $(k+1)$ appears in both terms. Carefully distribute any multiplications when working through brackets and parentheses.

Ultimately, this manipulation leads us to realize the structure of our initial formula expands correctly, reflecting it accurately for $k+1$. Good algebraic manipulation can make complex proofs like these more approachable. It allows you to see how individual problem pieces fit neatly together into a complete, final picture.

Problem 61

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