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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
Chapter6: Fractions
Section6.4: Least Common Denominators
Problem 13E
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Please help me with these two questions. I am having trouble understanding what to do

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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
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