Problem 88
Question
Find the sum. $$\sum_{i=1}^{6}(3 i-1)$$
Step-by-Step Solution
Verified Answer
The sum of the series is 57.
1Step 1: Calculate a_1 and a_n
To begin with, calculate the first term (a_1 = 3*1 - 1 = 2) and the last term (a_n = a_6 = 3*6 - 1 = 17).
2Step 2: Insert a_1 and a_n into the Sum formula
We will now use the sum of arithmetic series formula which is \(S_n = \frac{n}{2}(a_1 + a_n)\). Here, \(n = 6\), \(a_1 = 2\) and \(a_n = a_6 = 17\). Insert these into the formula to calculate the sum.
3Step 3: Compute the sum
By substituting the values, you obtain \(S_n = \frac{6}{2}(2 + 17)\), which simplifies to \(S_n = 3*(19)\ = 57\). This is the sum of the series.
Key Concepts
Sum FormulaSequence TermsArithmetic Progression
Sum Formula
In arithmetic series, a common way to find the sum is to use the sum formula. This formula is handy for quickly calculating the total of a sequence without adding each term individually. The sum formula for an arithmetic series is \( S_n = \frac{n}{2}(a_1 + a_n) \), where:
- Sn is the sum of the first n terms in the series
- n is the number of terms
- a1 is the first term
- an is the last term
Sequence Terms
In any sequence, understanding the terms is crucial. A sequence is composed of individual elements, often generated by a specific rule or formula. Each element in a sequence is referred to as a "term."
For the arithmetic sequence \( \sum_{i=1}^{6}(3i-1) \), each term is calculated by substituting values of i from 1 to 6 into the expression \(3i-1\).
This gives:
For the arithmetic sequence \( \sum_{i=1}^{6}(3i-1) \), each term is calculated by substituting values of i from 1 to 6 into the expression \(3i-1\).
This gives:
- a1 = 2 (when i=1)
- a2 = 5 (when i=2)
- a3 = 8 (when i=3)
- a4 = 11 (when i=4)
- a5 = 14 (when i=5)
- a6 = 17 (when i=6)
Arithmetic Progression
Arithmetic progression means that each term after the first is found by adding a constant difference to the previous term. This constant difference is called the "common difference."
For the sequence \( \sum_{i=1}^{6}(3i-1) \), you find that each term increases by 3 (i.e., 5-2 = 3, 8-5 = 3, and so on).
This consistent step is what defines an arithmetic progression.
For the sequence \( \sum_{i=1}^{6}(3i-1) \), you find that each term increases by 3 (i.e., 5-2 = 3, 8-5 = 3, and so on).
This consistent step is what defines an arithmetic progression.
- First term (\(a_1\)): 2
- Common difference (\(d\)): 3
- Next term (\(a_2\)): 5
Other exercises in this chapter
Problem 88
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Expand the expression in the difference quotient and simplify. \(\frac{f(x+h)-f(x)}{h}, h \neq 0\) \(f(x)=x^{8}\)
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Determine whether the statement is true or false. Justify your answer. Given the \(n\) th term and the common difference of an arithmetic sequence, it is possib
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Explain in your own words the meaning of \(_{n} P_{r}\).
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