Answer: Find the mistakes in the proof fragments in 3335

Chapter 5, Problem 33

(choose chapter or problem)

Theorem: For any integer n ≥ 1,

\(1^{2}+2^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}\).

“Proof (by mathematical induction): Certainly the theorem is true for n = 1 because \(1^{2}\) = 1 and \(\frac{1(1+1)(2 \cdot 1+1)}{6}=1\) So the basis step is true. For the inductive step, suppose that for some integer k ≥ 1, \(k^{2}=\frac{k(k+1)(2 k+1)}{6}\) . We must show that \((k+1)^{2}=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}\).”

Text Transcription:

1^2 + 2^2 + cdots + n^2 = n(n+1)(2 n+1) / 6

1^2

1(1+1)(2 cdot 1+1) / 6 = 1

k^2 = k(k+1)(2 k+1) / 6

(k+1)^2 = (k+1)((k+1)+1)(2(k+1)+1) / 6