Explanation Given information: ∑ k = 3 n ( k 2 − 2 k ) Concept used: Theorem of polynomial order. Calculation: ∑ k = 3 n ( k 2 − 2 k ) = ∑ k = 3 n k 2 − 2 ∑ k = 3 n k = 3 2 + 4 2 + 5 2 + ..... + n 2 − 2 [ 3 + 4 + 5 + .... + n ] = ( 1 2 + 2 2 + 3 3 + ..... n 2 − 1 2 − 2 2 ) − 2 ( 1 + 2 + 3 + ...... + n − 1 − 2 ) = ( 1 2 + 2 2 + 3 2 + ...... + n 2 ) − 5 − 2 ( 1 + 2 + 3 + ..... + n ) + 6 = ( 1 2 + 2 2 + 3 2 + ...... + n 2 ) − 2 ( 1 + 2 + 3 + ....

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ISBN: 9780357540244

Author: EPP

Publisher: CENGAGE L

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Chapter 11.2, Problem 47ES

To determine

To prove:

∑k=3n(k2−2k) is Θ(n3)

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Chapter 11.2, Problem 46ES

Chapter 11.2, Problem 48ES

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