Prove Sum Of Squares Using Mathematical Induction

by Alex Johnson 50 views

Introduction to Mathematical Induction

Mathematical induction is a powerful proof technique used in mathematics to establish that a particular statement or formula holds true for all positive integers. It's like a domino effect; if you can knock over the first domino and show that each domino will knock over the next one, then you know all the dominoes will fall. The principle of mathematical induction consists of two main steps: the base case and the inductive step. The base case involves verifying the statement for the smallest positive integer, usually n=1. The inductive step requires assuming the statement is true for an arbitrary positive integer, say 'k' (this is the inductive hypothesis), and then proving that the statement must also be true for the next integer, 'k+1'. By successfully completing both these steps, we can confidently conclude that the statement is valid for all positive integers.

The Statement to Prove: Sum of Squares

We are tasked with proving the following statement using mathematical induction: for any positive integer n, the sum of the squares of the first n positive integers is given by the formula:

12+22+32+⋯+n2=n(n+1)(2n+1)61^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2 n+1)}{6}

This formula tells us how to calculate the sum of the squares of consecutive numbers without having to add them all up individually. For instance, to find the sum of the squares of the first 5 positive integers (1² + 2² + 3² + 4² + 5²), we could use the formula directly: 5(5+1)(2∗5+1)6=5(6)(11)6=55\frac{5(5+1)(2*5+1)}{6} = \frac{5(6)(11)}{6} = 55. The beauty of mathematical induction is that it provides a rigorous way to prove that this formula works for every positive integer n.

Step 1: The Base Case

The first crucial step in mathematical induction is establishing the base case. This involves proving that the statement holds true for the smallest possible positive integer, which is n=1. Let's substitute n=1 into both sides of our formula:

  • Left-hand side (LHS): The sum of the squares of the first 1 positive integer is simply 12=11^2 = 1.
  • Right-hand side (RHS): Using the formula, we get 1(1+1)(2∗1+1)6=1(2)(3)6=66=1\frac{1(1+1)(2*1+1)}{6} = \frac{1(2)(3)}{6} = \frac{6}{6} = 1.

Since the LHS (1) equals the RHS (1), the statement is true for n=1. This successfully completes our base case, giving us the initial confirmation that our formula is on the right track.

Step 2: The Inductive Step

Now, we move on to the inductive step. This is where the real power of induction comes into play. We begin by making an assumption, known as the inductive hypothesis. We assume that the statement is true for some arbitrary positive integer k. That is, we assume:

12+22+32+⋯+k2=k(k+1)(2k+1)61^2+2^2+3^2+\cdots+k^2=\frac{k(k+1)(2 k+1)}{6}

Our goal is to use this assumption to prove that the statement must also be true for the next integer, k+1. In other words, we need to show that:

12+22+32+⋯+k2+(k+1)2=(k+1)((k+1)+1)(2(k+1)+1)61^2+2^2+3^2+\cdots+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}

Let's simplify the RHS for n=k+1 to see what we are aiming for:

(k+1)(k+2)(2k+2+1)6=(k+1)(k+2)(2k+3)6\frac{(k+1)(k+2)(2k+2+1)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}

Now, let's start with the LHS of the statement for n=k+1 and use our inductive hypothesis. The LHS is the sum of the squares up to k2k^2, plus the next term (k+1)2(k+1)^2:

LHS=(12+22+32+⋯+k2)+(k+1)2LHS = (1^2+2^2+3^2+\cdots+k^2) + (k+1)^2

According to our inductive hypothesis, we can replace the sum of the first k squares with the formula:

LHS=k(k+1)(2k+1)6+(k+1)2LHS = \frac{k(k+1)(2 k+1)}{6} + (k+1)^2

Our task now is to manipulate this expression algebraically to show that it is equal to the target formula for n=k+1, which is (k+1)(k+2)(2k+3)6\frac{(k+1)(k+2)(2k+3)}{6}.

Let's find a common denominator, which is 6:

LHS=k(k+1)(2k+1)6+6(k+1)26LHS = \frac{k(k+1)(2 k+1)}{6} + \frac{6(k+1)^2}{6}

Now, we can combine the fractions:

LHS=k(k+1)(2k+1)+6(k+1)26LHS = \frac{k(k+1)(2 k+1) + 6(k+1)^2}{6}

We can see a common factor of (k+1)(k+1) in the numerator. Let's factor it out:

LHS=(k+1)[k(2k+1)+6(k+1)]6LHS = \frac{(k+1) [k(2 k+1) + 6(k+1)]}{6}

Now, let's expand and simplify the expression inside the square brackets:

k(2k+1)+6(k+1)=2k2+k+6k+6=2k2+7k+6k(2 k+1) + 6(k+1) = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6

So, our expression for the LHS becomes:

LHS=(k+1)(2k2+7k+6)6LHS = \frac{(k+1)(2k^2 + 7k + 6)}{6}

Our final step in the inductive step is to factor the quadratic expression 2k2+7k+62k^2 + 7k + 6. We are looking for two numbers that multiply to 2∗6=122*6=12 and add up to 7. These numbers are 3 and 4. So we can rewrite the quadratic as:

2k2+7k+6=2k2+4k+3k+62k^2 + 7k + 6 = 2k^2 + 4k + 3k + 6

Factor by grouping:

2k(k+2)+3(k+2)=(2k+3)(k+2)2k(k+2) + 3(k+2) = (2k+3)(k+2)

Substituting this back into our LHS expression, we get:

LHS=(k+1)(k+2)(2k+3)6LHS = \frac{(k+1)(k+2)(2k+3)}{6}

This is exactly the formula we aimed to prove for n=k+1. We have successfully shown that if the statement is true for k, it is also true for k+1.

Conclusion

By successfully completing both the base case (proving the statement for n=1) and the inductive step (proving that if the statement holds for k, it also holds for k+1), we have rigorously proven, using the principle of mathematical induction, that the formula 12+22+32+⋯+n2=n(n+1)(2n+1)61^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2 n+1)}{6} is true for all positive integers n.

For further exploration into the fascinating world of proofs and number theory, you might find the resources at Wolfram MathWorld very informative. You can also check out Khan Academy for more lessons on mathematical induction.