Problem 6 - A Gentle Introduction to the Art of Mathematics

Question

Prove \(\sum_{j=0}^{n}(4 j+1)=2 n^{2}+3 n+1\) for all integers \(n \geq 0\).

Step-by-Step Solution

Answer

The sum formula is proven true for all integers \( n \geq 0 \) using mathematical induction.

Step 1: Base Case

First, verify the equation for the base case, when \( n = 0 \). Substituting \( n = 0 \) into the left-hand side of the equation, we get:\[ \sum_{j=0}^{0}(4j+1) = 4(0)+1 = 1 \]Substituting \( n = 0 \) into the right-hand side of the equation, we get:\[ 2(0)^2 + 3(0) + 1 = 1 \]Both sides are equal, so the base case holds.

Step 2: Inductive Hypothesis

Assume that the formula is true for some integer \( k \), i.e., \[ \sum_{j=0}^{k}(4j+1)=2k^{2}+3k+1 \]This is our induction hypothesis.

Step 3: Inductive Step

We need to prove that the formula holds for \( k + 1 \). Consider the sum from \( j = 0 \) to \( k + 1 \):\[ \sum_{j=0}^{k+1}(4j+1) = \sum_{j=0}^{k}(4j+1) + (4(k+1)+1) \]Using the inductive hypothesis:\[ = (2k^{2} + 3k + 1) + (4k + 4 + 1) \]Simplify the expression:\[ = 2k^{2} + 3k + 1 + 4k + 5 \]\[ = 2k^{2} + 7k + 6 \]

Step 4: Rewrite and Simplify

We need to show that this can be rewritten in the desired form:\[ 2(k+1)^{2} + 3(k+1) + 1 \]Expand the right-hand side:\[ = 2(k^2 + 2k + 1) + 3k + 3 + 1 \]\[ = 2k^{2} + 4k + 2 + 3k + 4 \]\[ = 2k^{2} + 7k + 6 \]Thus, both sides are equal, verifying the inductive step.

Step 5: Conclusion

Since both the base case and inductive step have been proven, by mathematical induction, the formula:\[ \sum_{j=0}^{n}(4j+1)=2n^{2}+3n+1 \]is true for all integers \( n \geq 0 \).

Key Concepts

Inductive Hypothesis

Understanding the inductive hypothesis is essential in any mathematical induction proof. It acts as the assumption step.
In our problem, the inductive hypothesis assumes that the given formula holds for some integer value, typically designated as \(k\).

For this exercise, the inductive hypothesis is: \[\sum_{j=0}^{k}(4j+1)=2k^{2}+3k+1\] In simpler terms, we are assuming the formula works perfectly for \(k\) without any errors.
This hypothesis is a critical bridge between the base case and proving the next step.

Base Case

The base case is our starting point. It's where we confirm that the given formula works for the smallest possible value of \(n\).

In this exercise, we test the value \(n=0\).
Substituting \(n=0\) in the left-hand side (LHS) and right-hand side (RHS) of our given formula:

• \[ LHS: \sum_{j=0}^{0}(4j+1) = 4(0)+1 = 1 \]
• \[ RHS: 2(0)^{2} + 3(0) + 1 = 1 \]

Since both sides are equal at \(n=0\), we can establish that the base case holds true.

Inductive Step

The inductive step is where we use our hypothesis to prove that the formula holds for \(k+1\).

We need to consider the sum for \(n=k+1\): \[\sum_{j=0}^{k+1}(4j+1) = \sum_{j=0}^{k}(4j+1) + (4(k+1)+1) \]Using the inductive hypothesis:\[\sum_{j=0}^{k}(4j+1) = 2k^{2} + 3k + 1\]
thus, \[\sum_{j=0}^{k+1}(4j+1) = (2k^{2} + 3k + 1) + (4k + 4 + 1) \] Simplifying, we get:

• \[ = 2k^{2} + 3k + 1 + 4k + 5 \]
• \[ = 2k^{2} + 7k + 6 \]


We rewrite the right-hand side to verify it's in the form we need.

Expand \(2(k+1)^2 + 3(k+1) + 1\): \[=2(k^2 + 2k + 1) + 3k + 3 + 1\]Simplifying the right-hand side further:\[=2k^2 + 4k + 2 + 3k + 3 + 1 = 2k^2 + 7k + 6\]

Since both sides match, we complete the inductive proof!