**Question 9** Use the drop-down menu to complete the proof of the following statement: \[ 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} \] for \( n = 1, 2, \ldots \). (Proof: Base Step) Assume \( P(1) \) is true: \[ 1^2 = \frac{1 \cdot 2 \cdot 3}{6} \] which is true since [dropdown menu]. (Inductive Step) Assume \( P(k) \) is true. So \[ 1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6} \] is true. We must show that [dropdown menu] is true. Notice \[ 1^2 + 2^2 + \cdots + (k+1)^2 \] \[ = 1^2 + 2^2 + \cdots + [dropdown menu] + (k+1)^2 \] \[ = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \] \[ = [dropdown menu] \left( \frac{k(2k+1)}{6} + (k+1) \right) \] \[ = (k+1) \left( \frac{k(2k+1) + 6(k+1)}{6} \right) \] \[ = (k+1) \left( \frac{2k^2 + 7k + 6}{6} \right) \] \[ = (k+1) \left( \frac{(k+2)(2k+3)}{6} \right) \] \[ = \frac{(k+1)(k+2)(2k+3)}{6} \] Therefore, \( P(k+1) \) is [dropdown menu].
**Question 9** Use the drop-down menu to complete the proof of the following statement: \[ 1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6} \] for \( n = 1, 2, \ldots \). (Proof: Base Step) Assume \( P(1) \) is true: \[ 1^2 = \frac{1 \cdot 2 \cdot 3}{6} \] which is true since [dropdown menu]. (Inductive Step) Assume \( P(k) \) is true. So \[ 1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6} \] is true. We must show that [dropdown menu] is true. Notice \[ 1^2 + 2^2 + \cdots + (k+1)^2 \] \[ = 1^2 + 2^2 + \cdots + [dropdown menu] + (k+1)^2 \] \[ = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \] \[ = [dropdown menu] \left( \frac{k(2k+1)}{6} + (k+1) \right) \] \[ = (k+1) \left( \frac{k(2k+1) + 6(k+1)}{6} \right) \] \[ = (k+1) \left( \frac{2k^2 + 7k + 6}{6} \right) \] \[ = (k+1) \left( \frac{(k+2)(2k+3)}{6} \right) \] \[ = \frac{(k+1)(k+2)(2k+3)}{6} \] Therefore, \( P(k+1) \) is [dropdown menu].
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
![**Question 9**
Use the drop-down menu to complete the proof of the following statement:
\[
1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}
\]
for \( n = 1, 2, \ldots \).
(Proof: Base Step) Assume \( P(1) \) is true:
\[
1^2 = \frac{1 \cdot 2 \cdot 3}{6}
\]
which is true since [dropdown menu].
(Inductive Step) Assume \( P(k) \) is true. So
\[
1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}
\]
is true.
We must show that [dropdown menu] is true. Notice
\[
1^2 + 2^2 + \cdots + (k+1)^2
\]
\[
= 1^2 + 2^2 + \cdots + [dropdown menu] + (k+1)^2
\]
\[
= \frac{k(k+1)(2k+1)}{6} + (k+1)^2
\]
\[
= [dropdown menu] \left( \frac{k(2k+1)}{6} + (k+1) \right)
\]
\[
= (k+1) \left( \frac{k(2k+1) + 6(k+1)}{6} \right)
\]
\[
= (k+1) \left( \frac{2k^2 + 7k + 6}{6} \right)
\]
\[
= (k+1) \left( \frac{(k+2)(2k+3)}{6} \right)
\]
\[
= \frac{(k+1)(k+2)(2k+3)}{6}
\]
Therefore, \( P(k+1) \) is [dropdown menu].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2F82d4da3a-1c23-46ee-a4ce-4180c9026b41%2Fp81v68r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 9**
Use the drop-down menu to complete the proof of the following statement:
\[
1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}
\]
for \( n = 1, 2, \ldots \).
(Proof: Base Step) Assume \( P(1) \) is true:
\[
1^2 = \frac{1 \cdot 2 \cdot 3}{6}
\]
which is true since [dropdown menu].
(Inductive Step) Assume \( P(k) \) is true. So
\[
1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}
\]
is true.
We must show that [dropdown menu] is true. Notice
\[
1^2 + 2^2 + \cdots + (k+1)^2
\]
\[
= 1^2 + 2^2 + \cdots + [dropdown menu] + (k+1)^2
\]
\[
= \frac{k(k+1)(2k+1)}{6} + (k+1)^2
\]
\[
= [dropdown menu] \left( \frac{k(2k+1)}{6} + (k+1) \right)
\]
\[
= (k+1) \left( \frac{k(2k+1) + 6(k+1)}{6} \right)
\]
\[
= (k+1) \left( \frac{2k^2 + 7k + 6}{6} \right)
\]
\[
= (k+1) \left( \frac{(k+2)(2k+3)}{6} \right)
\]
\[
= \frac{(k+1)(k+2)(2k+3)}{6}
\]
Therefore, \( P(k+1) \) is [dropdown menu].
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