Consider the sum (3)² + (8)² + (13)2 + · . + (15n – 2)2. | m 1 1 m(т + 1)] and k2 = (m(m + 1)(2m + 1)] to prove that k=1 Use identities k (3)² + (8)² + (13)² +· …+(15n – 2)² =;(n) (450n² + 45n – 11). 150n² - ...
Consider the sum (3)² + (8)² + (13)2 + · . + (15n – 2)2. | m 1 1 m(т + 1)] and k2 = (m(m + 1)(2m + 1)] to prove that k=1 Use identities k (3)² + (8)² + (13)² +· …+(15n – 2)² =;(n) (450n² + 45n – 11). 150n² - ...
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Consider the sum (3)² + (8)² + (13)² + · · + (15n – 2)² .
1
[m(m + 1)] and
— m(т + 1)(2т + 1)] to prove that
Use identities
k=1
(3)² + (8)² + (13)² + ..+ (15n – 2)² = (n) (450n2 + 45n – 11).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7614b3cb-8aec-4087-8c86-043f02640bd9%2F1b3f5f42-d377-4aa1-992e-1237de7980a2%2Fec18q0f_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the sum (3)² + (8)² + (13)² + · · + (15n – 2)² .
1
[m(m + 1)] and
— m(т + 1)(2т + 1)] to prove that
Use identities
k=1
(3)² + (8)² + (13)² + ..+ (15n – 2)² = (n) (450n2 + 45n – 11).
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