Explanation Given information: lim n → ∞ ∑ k = 1 n ( k 2 + 1 ) n 3 Concept and Formula Used: ∑ k = 1 n ( k 2 ) = n ( n + 1 ) ( 2 n + 1 ) 6 , ∑ k = 1 n ( c ) = c n where c is constant Calculation: The given summation is lim n → ∞ ∑ k = 1 n ( k 2 + 1 ) n 3 lim n → ∞ ∑ k = 1 n ( k 2 + 1 ) n 3 = lim n → ∞ 1 n 3 ( ∑ k = 1 n ( k 2 ) + ∑ k = 1 n ( 1 ) ) Using ∑ k = 1 n ( k 2 ) = n ( n + 1 ) ( 2 n + 1 ) 6 , ∑ k = 1 n ( c ) = c n where c is constant lim n → ∞ ∑ k = 1 n ( k 2 + 1 ) n 3 = lim n → ∞ 1 n 3 ( n ( n + 1 ) ( 2 n + 1 ) 6 + n ) lim n → ∞ ∑ k = 1 n

5th Edition

ISBN: 9781429241861

Author: Laura Taalman, Peter Kohn

Publisher: W. H. Freeman

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Sequence and Series

A sequence is defined as a collection of things. Series is defined to sum the things one by one in the sequence. It was invented by German mathematician Carl Friedrich Gauss.

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Chapter 4, Problem 5SC

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To find: limn→∞∑k=1n( k 2 +1)n3

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Chapter 4, Problem 4SC

Chapter 4, Problem 6SC

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Chapter 4 Solutions

Calculus

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lim n → ∞ ∑ k = 1 n ( k 2 + 1 ) n 3

Solution Summary: The author calculates the summation of lundersetntoinfty

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To find: limn→∞∑k=1n( k 2 +1)n3

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Chapter 4 Solutions