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Chapter 7: Problem 8

Evaluate the arithmetic series. $$ \sum_{m=1}^{75}(2+3 m) $$

Short Answer

Expert verified
The sum of the given arithmetic series, \(\sum_{m=1}^{75}(2+3 m)\), is 8700, using the formula for the sum of an arithmetic series \(S_n = \frac{n}{2}(a + l)\), where \(n = 75\), \(a = 5\), and \(l = 227\).

Step by step solution

01

Find the number of terms (n)

From the given series, we can see that it starts from m=1 and goes up to m=75. Hence, the number of terms in the series is \(n = 75\).
02

Identify the common difference (d)

In the series, we can see that the general term is given by \(T_m = 2+ 3m\). If we find the difference between subsequent terms, we can find the common difference (d). If we define \(T_2\) as the second term and \(T_1\) as the first term, we can find the common difference as, \(d = T_2 - T_1 \) which is equivalent to \(d = (2+3(2)) - (2+3(1)) = 5 - 2 = 3\). So, the common difference (d) in the given series is 3.
03

Find the first term (a) and the last term (l)

Now, we identify the first term (a) and the last term (l) of the given series. The first term is given by \(a = 2 + 3(1) = 5\). The last term is given by \(l = 2 + 3(75) = 2 + 225 = 227\).
04

Use the formula for the sum of an arithmetic series

The sum (S) of an arithmetic series can be calculated using the following formula: \[S_n = \frac{n}{2}(a + l)\] Where, \(S_n\) is the sum of the series, \(n\) is the number of terms, a is the first term, and l is the last term. Substitute the values we've found into the formula: \[S_{75} = \frac{75}{2}(5 + 227)\]
05

Calculate the sum of the series

Now, to find the sum of the series, perform the operations in the formula: \[S_{75} = \frac{75}{2}(232) = 37.5 * 232 = 8700\] So, the sum of the given arithmetic series is 8700.

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Most popular questions from this chapter

Explain why $$ \sum_{m=1}^{999}\left(m^{5}-2 m+7\right)=\sum_{k=1}^{999}\left(k^{5}-2 k+7\right) . $$

Show that an infinite sequence \(a_{1}, a_{2}, a_{3}, \ldots\) is an arithmetic sequence if and only if there is a linear function \(f\) such that $$ a_{n}=f(n) $$ for every positive integer \(n\).

Show that $$ \sqrt{n^{2}+n}-n=\frac{1}{\sqrt{1+\frac{1}{n}}+1} $$ [Hint: Multiply the expression \(\sqrt{n^{2}+n}-n\) by \(\left(\sqrt{n^{2}+n}+n\right) /\left(\sqrt{n^{2}+n}+n\right) .\) Then factor \(n\) out of the numerator and denominator of the resulting expression.] [This identity was used in Example 1.]

Show that $$ \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<\ln n $$ for every integer \(n \geq 2\). [Hint: Draw the graph of the curve \(y=\frac{1}{x}\) in the \(x y\) plane. Think of \(\ln n\) as the area under part of this curve. Draw appropriate rectangles under the curve.]

Express $$ 8.237545454 \ldots $$ as a fraction; here the digits 54 repeat forever.
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