This exercise uses one or more of the following results from Section 5.2. Exercise 5.2.10: For every integer n 2 1, 12 + 22 + + 02 = n(n + 1)(2n + 1) 6 n(n + 1) Exercise 5.2.11: For every integer n 2 1, 13 + 23 + -. + n3 = 2 Theorem 5.2.1: For every integer n 2 1, 1 + 2 +- +n= n(n + 1) Theorem 5.2.2: For any real number rexcept 1, and any integer n 2 0, - --1 r- 1 -0 Theorem on Polynomial Orders: If m is any integer with m 2 0 and ao, a, a,., am are real numbers with a, > 0, then anm + am,nm -1 +... + a,n + a, is O(n"). Prove the following statement. 5 + 10 + 15 + 20 + + 5n is O(n2). Proof: We get )(+ +2+3*** 5 + 10 + 15 + 20 + + 5n = = 51 by-Select--- %3! which is a polynomial of degree 1. So, by 5 + 10 + 15 + 20 + 25 + - + 5n is 0 n --Select--
This exercise uses one or more of the following results from Section 5.2. Exercise 5.2.10: For every integer n 2 1, 12 + 22 + + 02 = n(n + 1)(2n + 1) 6 n(n + 1) Exercise 5.2.11: For every integer n 2 1, 13 + 23 + -. + n3 = 2 Theorem 5.2.1: For every integer n 2 1, 1 + 2 +- +n= n(n + 1) Theorem 5.2.2: For any real number rexcept 1, and any integer n 2 0, - --1 r- 1 -0 Theorem on Polynomial Orders: If m is any integer with m 2 0 and ao, a, a,., am are real numbers with a, > 0, then anm + am,nm -1 +... + a,n + a, is O(n"). Prove the following statement. 5 + 10 + 15 + 20 + + 5n is O(n2). Proof: We get )(+ +2+3*** 5 + 10 + 15 + 20 + + 5n = = 51 by-Select--- %3! which is a polynomial of degree 1. So, by 5 + 10 + 15 + 20 + 25 + - + 5n is 0 n --Select--
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:This exercise uses one or more of the following results from Section 5.2.
Exercise 5.2.10: For every integer n 2 1, 12 + 22 +
+ 0? = n(n + 1)(2n + 1)
6
n(n + 1)
Exercise 5.2.11: For every integer n 2 1, 13 + 23 + -.. + n3 =
2
Theorem 5.2.1: For every integer n 2 1, 1 +2+... + 0 = n(h + 1)
Theorem 5.2.2: For any real number rexcept 1, and any integer n 2 0, = *-1
r- 1
-0
Theorem on Polynomial Orders: If m is any integer with m 2 0 and ao, a, a,.., am are real numbers with a > 0, then anm + am,nm -1+... + a,n + a, is O(n")
Prove the following statement.
5 + 10 + 15 + 20 +... + 5n is O(n2).
Proof: We get
)(+2+3+4+
+n)
5 + 10 + 15 + 20 + + 5n =
= 51
by -Select--
!3!
which is a polynomial of degree
1. So, by
D 5+ 10 + 15 + 20 + 25 +
+ 5n is 0n
--Select--
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