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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

Expert verified
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.

Step by step solution

01

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.
02

State the Induction Hypothesis

Assume that the given formula holds true for n = k, where k is an arbitrary positive integer: \[1^{2}+2^{2}+3^{2}+\ldots+k^{2} = \frac{1}{6}k(k+1)(2k+1).\]
03

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:
04

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: \[\frac{1}{6}(2k^2 + k) + \frac{6(k+1)}{6} = \frac{1}{6}(2k^2 + k + 6k + 6) = \frac{1}{6}(2k^2 + 7k + 6).\]
05

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.
06

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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