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Ch. 5 - Integrals
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 5, Problem 5.PE.3c

           10           10
Suppose that Σ aₖ = -2 and Σ bₖ = 25. Find the value of
           k = 1          k = 1


  10
c. Σ (aₖ + bₖ - 1)
  k = 1

Verified step by step guidance
1
Identify the given sums: \( \sum_{k=1}^{10} a_k = -2 \) and \( \sum_{k=1}^{10} b_k = 25 \).
Write the expression to find: \( \sum_{k=1}^{10} (a_k + b_k - 1) \).
Use the property of summation that allows splitting sums: \( \sum_{k=1}^{10} (a_k + b_k - 1) = \sum_{k=1}^{10} a_k + \sum_{k=1}^{10} b_k - \sum_{k=1}^{10} 1 \).
Substitute the known sums into the expression: \( = (-2) + 25 - \sum_{k=1}^{10} 1 \).
Calculate \( \sum_{k=1}^{10} 1 \) as the sum of ten ones, which equals 10, then combine all terms to express the final sum.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Properties of Summation

Summation properties allow us to break down or combine sums term-by-term. For example, the sum of a sum is the sum of the sums: Σ(aₖ + bₖ) = Σaₖ + Σbₖ. This linearity helps simplify complex summations by separating or combining known series.
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Constant Term in Summation

When summing a constant over n terms, the result is the constant multiplied by n: Σc = n * c. This is useful when the summation includes a constant term added or subtracted from each element, allowing easy calculation of that part of the sum.
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Given Summation Values

Knowing the values of Σaₖ and Σbₖ provides a foundation to find related sums. By substituting these known sums into expressions involving aₖ and bₖ, we can compute new summations without evaluating each term individually.
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