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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 Step 3 Step 4 We observe that all the integers from 1 to n are at most n, hence 1k + 2k + +nk ≤nk + nk +...+nk. By combining our calculations, we get 1k + 2k + +nk ≤nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k +2k++n* 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 + nk++nk=n·nk. Using a law of exponentiation, we simplify nnk=nk+1, Using a law of exponentiation, we simplify n.nk=nk+1, By combining our calculations, we get 1k+2k +...+nk ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k + 2k + +n* to be O(n+1). We can simplify the sum by adding the exponents: nk + nk + + n² = nk+k+_+k = nnk We observe that all the integers from 1 to n are at most n, hence 1k + 2k++n* ≥nk + nk +...+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 Step 3 Step 4 We observe that all the integers from 1 to n are at most n, hence 1k + 2k + +nk ≤nk + nk +...+nk. By combining our calculations, we get 1k + 2k + +nk ≤nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k +2k++n* 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 + nk++nk=n·nk. Using a law of exponentiation, we simplify nnk=nk+1, Using a law of exponentiation, we simplify n.nk=nk+1, By combining our calculations, we get 1k+2k +...+nk ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k + 2k + +n* to be O(n+1). We can simplify the sum by adding the exponents: nk + nk + + n² = nk+k+_+k = nnk We observe that all the integers from 1 to n are at most n, hence 1k + 2k++n* ≥nk + nk +...+nk.

Advanced Engineering Mathematics
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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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 Step 3 Step 4 We observe that all the integers from 1 to n are at most n, hence 1k + 2k + +nk ≤nk + nk +...+nk. By combining our calculations, we get 1k +2k +...+nk ≤nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k + 2k +...+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 + nk +...+ n² = n・nk. Using a law of exponentiation, we simplify nnk=nk+1. Using a law of exponentiation, we simplify n.nk = nk+1. By combining our calculations, we get 1k + 2k + ... + n² ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k + 2k +...+nk to be O(n+1). We can simplify the sum by adding the exponents: nk + nk +...+ n² = nk+k+.+k = nnk We observe that all the integers from 1 to n are at most n, hence 1k + 2k + +nk ≥nk + nk + ... + 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 Step 3 Step 4 We observe that all the integers from 1 to n are at most n, hence 1k + 2k + +nk ≤nk + nk +...+nk. By combining our calculations, we get 1k +2k +...+nk ≤nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k + 2k +...+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 + nk +...+ n² = n・nk. Using a law of exponentiation, we simplify nnk=nk+1. Using a law of exponentiation, we simplify n.nk = nk+1. By combining our calculations, we get 1k + 2k + ... + n² ≥nk+1 for all n ≥ 1. We have verified the definition of what it means for 1k + 2k +...+nk to be O(n+1). We can simplify the sum by adding the exponents: nk + nk +...+ n² = nk+k+.+k = nnk We observe that all the integers from 1 to n are at most n, hence 1k + 2k + +nk ≥nk + nk + ... + nk.
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