Chapter 1: Chapter 1

Problem 18

Prove by mathematical induction $1^{2}+2^{2}+3^{2}+\ldots+n^{2}=(1 / 6) n(n+1)(2 n+1)$.

Short Answer

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By mathematical induction, we proved that the formula for the sum of the first n squares is true for all positive integers n: $1^2+2^2+3^2+\ldots+n^2 = \frac{1}{6} n(n+1)(2n+1)$. The proof consists of the following steps: 1. Verify the base case (n=1). 2. State the induction hypothesis for n=k. 3. Prove the induction step for n=k+1. 4. Simplify the RHS of the equation. 5. Relate the simplified RHS to the desired form. 6. Conclude that the formula holds true for all positive integers n.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Verify the Base Case

Start by verifying the formula for the base case, n = 1. The left-hand side (LHS) of the equation is the sum of the first n squares, and the right-hand side (RHS) is the given formula. For n = 1: LHS: $1^2 = 1$ RHS: $\frac{1}{6}(1)(1+1)(2(1)+1) = \frac{1}{6}(1)(2)(3) = 1$ Since the LHS and RHS are equal for n = 1, the base case is true.

Step 2: State the Induction Hypothesis

Assume that the given formula holds true for n = k, where k is an arbitrary positive integer: \[1^{2}+2^{2}+3^{2}+\ldots+k^{2} = \frac{1}{6}k(k+1)(2k+1).\]

Step 3: Prove the Induction Step

We have to prove that the formula also holds true for n = k+1. Consider the sum of the first (k+1) squares: $1^2+2^2+3^2+\ldots+k^2+(k+1)^2$ Now, we can apply the induction hypothesis to the sum of the first k squares: \[\biggl(1^{2}+2^{2}+3^{2}+\ldots+k^{2}\biggr) + (k+1)^{2} = \frac{1}{6}k(k+1)(2k+1) + (k+1)^2.\] Next, we need to simplify the RHS of the equation and try to bring it to the desired form for n = k+1:

Step 4: Simplify the RHS

To simplify the RHS, start by factoring out (k+1) from the two terms: \[\frac{1}{6}k(k+1)(2k+1) + (k+1)^2 = (k+1)\biggl(\frac{1}{6}k(2k+1) + (k+1)\biggr).\] Now, focus on simplifying the expression inside the parentheses: \[\frac{1}{6}k(2k+1) + (k+1) = \frac{1}{6}(2k^2 + k) + (k+1).\] Observe that we can rewrite the (k+1) term as $\frac{6(k+1)}{6}$, then combine the terms under a common denominator: \[\frac{1}{6}(2k^2 + k) + \frac{6(k+1)}{6} = \frac{1}{6}(2k^2 + k + 6k + 6) = \frac{1}{6}(2k^2 + 7k + 6).\]

Step 5: Relate the Simplified RHS to the Desired Form

Notice that we can factor the simplified RHS expression: \[\frac{1}{6}(2k^2 + 7k + 6) = \frac{1}{6}(k+1)((2k+1)+(k+2)).\] Comparing this to the desired form for n = k+1: \[\frac{1}{6}(k+1)(2(k+1)+1)(k+1+1) = \frac{1}{6}(k+1)(2k+3)(k+2).\] Since the simplified RHS is equal to the desired form, we have proven that the given formula holds true for n=k+1, thus completing the induction step.

Step 6: Conclusion

By induction, we have proven that the given formula for the sum of the first n squares is true for all positive integers n: \[1^{2}+2^{2}+3^{2}+\ldots+n^{2} = \frac{1}{6} n(n+1)(2n+1).\]

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Chapter 1: Chapter 1

Prove by mathematical induction $1^{2}+2^{2}+3^{2}+\ldots+n^{2}=(1 / 6) n(n+1)(2 n+1)$.

Short Answer

Step by step solution

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Step 1: Verify the Base Case

Step 2: State the Induction Hypothesis

Step 3: Prove the Induction Step

Step 4: Simplify the RHS

Step 5: Relate the Simplified RHS to the Desired Form

Step 6: Conclusion

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