Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Click and drag the steps that are required to prove 1k+ 2k +...+nk is O(nk + 1) to their corresponding step numbers. Step 1 Step 2 Using a law of exponentiation, we simplify n⋅nk=nk+1 Using a law of exponentiation, we simplify We observe that all the integers from 1 to n are at most n, hence 1+2++ n ≤ n* + n* + + n*. Step 3 We can simplify the sum by adding the exponents: nk+n++nk =nk+k+.+k=nk Step 4 We observe that all the integers from 1 to n are at most n, hence 1+2++ n ≥ n* + n* + ... + n*. By combining our calculations, we get 1 + 2k ++*** for all n ≥ 1. We have verified the definition of what it means for 1+2++nk to be O(n*+1). By combining our calculations, we get 1 +2*++n* s*+1 for all n ≥1. We have verified the definition of what it means for 1+2++nk to be O(n+1). Since there are exactly n equal terms in the sum, we can write the sum as a simple product: n+n++n* =n⋅nk.

Click and drag the steps that are required to prove 1k+ 2k +...+nk is O(nk + 1) to their corresponding step numbers. Step 1 Step 2 Using a law of exponentiation, we simplify n⋅nk=nk+1 Using a law of exponentiation, we simplify We observe that all the integers from 1 to n are at most n, hence 1+2++ n ≤ n* + n* + + n*. Step 3 We can simplify the sum by adding the exponents: nk+n++nk =nk+k+.+k=nk Step 4 We observe that all the integers from 1 to n are at most n, hence 1+2++ n ≥ n* + n* + ... + n*. By combining our calculations, we get 1 + 2k ++*** for all n ≥ 1. We have verified the definition of what it means for 1+2++nk to be O(n*+1). By combining our calculations, we get 1 +2*++n* s*+1 for all n ≥1. We have verified the definition of what it means for 1+2++nk to be O(n+1). Since there are exactly n equal terms in the sum, we can write the sum as a simple product: n+n++n* =n⋅nk.

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter5: Factoring Polynomials
Section5.7: Factoring Pattern For X2+bx+c, C Positive
Problem 5OE
See similar textbooks
icon
Related questions
Question

Please help me with these two questions. I am having trouble understanding what to do

Thank you 

Click and drag the steps that are required to prove 1k+ 2k +...+nk is O(nk + 1) to their corresponding step numbers. Step 1 Step 2 Using a law of exponentiation, we simplify n⋅nk=nk+1. Using a law of exponentiation, we simplify naknk+1 We observe that all the integers from 1 to n are at most n, hence 1k+2k++ n* ≤ n k +nk + ... + nk. Step 3 We can simplify the sum by adding the exponents: nk+n++ nk = nk+k+.+k = nrk. Step 4 We observe that all the integers from 1 to n are at most n, hence 1k+2k++nk ≥nk + n + ... + n*. By combining our calculations, we get 1 + 2k ++nk ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1+2++nk to be O(n+1). By combining our calculations, we get 1k + 2k + + n ≤n*+1 for all n≥1. We have verified the definition of what it means for 1+2++nk to be O(n+1). Since there are exactly n equal terms in the sum, we can write the sum as a simple product: nk+n++nk = n⋅nk.
Transcribed Image Text:Click and drag the steps that are required to prove 1k+ 2k +...+nk is O(nk + 1) to their corresponding step numbers. Step 1 Step 2 Using a law of exponentiation, we simplify n⋅nk=nk+1. Using a law of exponentiation, we simplify naknk+1 We observe that all the integers from 1 to n are at most n, hence 1k+2k++ n* ≤ n k +nk + ... + nk. Step 3 We can simplify the sum by adding the exponents: nk+n++ nk = nk+k+.+k = nrk. Step 4 We observe that all the integers from 1 to n are at most n, hence 1k+2k++nk ≥nk + n + ... + n*. By combining our calculations, we get 1 + 2k ++nk ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1+2++nk to be O(n+1). By combining our calculations, we get 1k + 2k + + n ≤n*+1 for all n≥1. We have verified the definition of what it means for 1+2++nk to be O(n+1). Since there are exactly n equal terms in the sum, we can write the sum as a simple product: nk+n++nk = n⋅nk.
For each of these functions, find the least integer n such that f(x) is O(x). If f(x) = 2x² + x³logx, then the least integer n is
Transcribed Image Text:For each of these functions, find the least integer n such that f(x) is O(x). If f(x) = 2x² + x³logx, then the least integer n is
Expert Solution
steps

Step by step

Solved in 3 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
Algebra
ISBN:
9780395977224
Author:
Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:
McDougal Littell
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
College Algebra
College Algebra
Algebra
ISBN:
9781305115545
Author:
James Stewart, Lothar Redlin, Saleem Watson
Publisher:
Cengage Learning
College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
Algebra
ISBN:
9781305652231
Author:
R. David Gustafson, Jeff Hughes
Publisher:
Cengage Learning
Glencoe Algebra 1, Student Edition, 9780079039897…
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill
SEE MORE TEXTBOOKS