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Chapter 14: Problem 41

Express each sum using summation notation. Use 1 as the lower limit of summation and i for the index of summation. $$1+3+5+\cdots+(2 n-1)$$

Short Answer

Expert verified
The sum \(1+3+5+\cdots+(2 n-1)\) in sigma notation with 1 as lower limit and i as index of summation can be written as \(\sum_{i=1}^{n}(2i-1)\).

Step by step solution

01

Identifying the first term and the common difference

Recognize that the series \(1+3+5+\cdots+(2 n-1)\) is an arithmetic series that begins with 1, increments by 2, and ends with the term \((2n-1)\). The first term a is 1, and the common difference d is 2.
02

Express the nth term in terms of i

The nth term of an arithmetic sequence can be found with the formula \(a + (n - 1) * d\), where a is the first term, n is the term number, and d is the common difference. In this case, the nth term of the sequence is given explicitly as \((2n-1)\). However, we want to find a general expression in terms of i, so we need to modify it. Using the formula and substituting the values a=1 and d=2, we get the nth term as \((2i-1)\).
03

Write in sigma notation

Finally, represent the series in sigma notation. Sigma notation provides a compact way to represent series. The notation \(\sum_{i=1}^{n}\), signifies a sum of terms. The variable i, is the index of summation; n is the upper limit of summation. The sum starts at i = 1 and ends at i = n. The term to be added, derived from the arithmetic series, is \((2i-1)\). Thus, the final sigma notation is \(\sum_{i=1}^{n}(2i-1)\).

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Most popular questions from this chapter

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. It makes a difference whether or not I use parentheses around the expression following the summation symbol, because the value of \(\sum_{i=1}^{6}(i+7)\) is \(92,\) but the value of \(\sum_{i=1}^{8} i+7\) is 43.

Each exercise involves observing a pattern in the expanded form of the binomial expression \((a+b)^{n}\).$$\begin{array}{l}(a+b)^{1}=a+b \\\\(a+b)^{2}=a^{2}+2 a b+b^{2} \\\\(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3} \\\\(a+b)^{4}=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\\\(a+b)^{5}=a^{5}+5 a^{4} b+10 a^{3} b^{2}+10 a^{2} b^{3}+5 a b^{4}+b^{5}\end{array}$$ Describe the pattern for the exponents on \(a\).

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. There's no end to the number of geometric sequences that I can generate whose first term is 5 if I pick nonzero numbers \(r\) and multiply 5 by each value of \(r\) repeatedly.

Rationalize the denominator: \(\frac{6}{\sqrt{3}-\sqrt{5}}\).

Explain how to find \(n !\) if \(n\) is a positive integer.
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