In Exercises sumfirst - sumlast, find the value of each sum using Definition sigmanotation.

\( \sum _{g = 4}^{9} (5g + 3)\)

\(\answer {213}\)

\( \sum _{k = 3}^{8} \frac {1}{k}\)

\(\answer {\frac {341}{280}}\)

\( \sum _{j = 0}^{5} 2^{j}\)

\(\answer {63}\)

\( \sum _{k = 0}^{2} (3k - 5)x^{k}\)

\(\answer {-5 - 2x + x^{2}}\)

\( \sum _{i = 1}^{4} \frac {1}{4}(i^{2} + 1)\)

\(\answer {\frac {17}{2}}\)

\( \sum _{n = 1}^{100} (-1)^{n}\)

\(\answer {0}\)

\( \sum _{n = 1}^{5} \frac {(n+1)!}{n!}\)

\(\answer {20}\)

\( \sum _{j = 1}^{3} \frac {5!}{j! \, (5-j)!}\)

\(\answer {25}\)

In Exercises writesumfirst - writesumlast, rewrite the sum using summation notation.

\(8 + 11 + 14 + 17 + 20\)

\(\displaystyle \sum _{k = 1}^{5} (3k + 5)\)
\(1 - 2 + 3 - 4 + 5 - 6 + 7 - 8\)

\(\displaystyle \sum _{k = 1}^{8} (-1)^{k - 1}k\)
\(x - \frac {x^{3}}{3} + \frac {x^{5}}{5} - \frac {x^{7}}{7}\)

\(\displaystyle \sum _{k = 1}^{4} (-1)^{k - 1} \frac {x^{2k - 1}}{2k - 1}\)
\(1 + 2 + 4 + \cdots + 2^{29}\)

\(\displaystyle \sum _{k = 1}^{30} 2^{k-1}\)
\(2 + \frac {3}{2} + \frac {4}{3} + \frac {5}{4} + \frac {6}{5}\)

\(\displaystyle \sum _{k = 1}^{5} \frac {k + 1}{k}\)
\(-\ln (3) + \ln (4) - \ln (5) + \cdots + \ln (20)\)

\(\displaystyle \sum _{k = 3}^{20} (-1)^{k} \ln (k)\)
\(1 - \frac {1}{4} + \frac {1}{9} - \frac {1}{16} + \frac {1}{25} - \frac {1}{36}\)

\(\displaystyle \sum _{k = 1}^{6} \frac {(-1)^{k - 1}}{k^{2}}\)
\(\frac {1}{2}(x - 5) + \frac {1}{4}(x - 5)^{2} + \frac {1}{6}(x - 5)^{3} + \frac {1}{8}(x - 5)^{4}\)

\(\displaystyle \sum _{k = 1}^{4} \frac {1}{2k}(x - 5)^{k}\)

In Exercises findsumformfirst - findsumformulalast, use the formulas in Equation arithgeosum to find the sum.

\( \sum _{n = 1}^{10} 5n+3\)

\(\answer {305}\)

\( \sum _{n = 1}^{20} 2n-1\)

\(\answer {400}\)

\( \sum _{k = 0}^{15} 3-k\)

\(\answer {-72}\)

\( \sum _{n = 1}^{10} \left (\frac {1}{2}\right )^{n}\)

\(\answer {\frac {1023}{1024}}\)

\( \sum _{n = 1}^{5} \left (\frac {3}{2}\right )^{n}\)

\(\answer {\frac {633}{32}}\)

\( \sum _{k = 0}^{5} 2\left (\frac {1}{4}\right )^{k}\)

\(\answer {\frac {1365}{512}}\)

\(1+4+7+ \ldots +295\)

\(\answer {14652}\)

\(4+2+0-2- \ldots - 146\)

\(\answer {-5396}\)

\(1+3+9+ \ldots + 2187\)

\(\answer {3280}\)

\(\frac {1}{2} + \frac {1}{4} + \frac {1}{8} + \ldots + \frac {1}{256}\)

\(\answer {\frac {255}{256}}\)

\(3 - \frac {3}{2} + \frac {3}{4} - \frac {3}{8}+- \dots +\frac {3}{256}\)

\(\answer {\frac {513}{256}}\)

\( \sum _{n = 1}^{10} -2n + \left (\frac {5}{3}\right )^{n}\)

\(\answer {\frac {17771050}{59049}}\)

In Exercises geoseriesexamplefirst - geoseriesexamplelast, use Theorem geoseries to find the sum of the given geometric series.

\( \sum _{n = 1}^{\infty } \left ( \frac {1}{2} \right )^{n-1}\)

\(\answer {2}\)

\( \sum _{n = 0}^{\infty } \frac {(-1)^n \, 3^{n-1}}{4^{n}}\)

\(\answer {\frac {4}{21}}\)

\( \sum _{m = 2}^{\infty } \frac {3}{2^{m-1}}\)

\(\answer {3}\)

\( \sum _{k =0}^{\infty } x^{k}\), \(|x|<1\)

\(\answer {\frac {1}{1-x}}\)

In Exercises dectofracfirst - dectofraclast, use Theorem geoseries to express each repeating decimal as a fraction of integers.

\(0.\overline {7}\)

\(\frac {7}{9}\)
\(0.\overline {13}\)

\(\frac {13}{99}\)
\(10.\overline {159}\)

\(\frac {3383}{333}\)
\(-5.8\overline {67}\)

\(-\frac {5809}{990}\)
In Exercises annuityfirst - annuitylast, use Equation fvannuity to compute the future value of the annuity with the given terms. In all cases, assume the payment is made monthly, the interest rate given is the annual rate, and interest is compounded monthly.
payments are $300, interest rate is 2.5%, term is 17 years.

\(\$76,\!163.67\)
payments are $50, interest rate is 1.0%, term is 30 years.

\(\$20,\!981.40\)
payments are $100, interest rate is 2.0%, term is 20 years

\(\$29,\!479.69\)
payments are $100, interest rate is 2.0%, term is 25 years

\(\$38,\!882.12\)
payments are $100, interest rate is 2.0%, term is 30 years

\(49,\!272.55\)
payments are $100, interest rate is 2.0%, term is 35 years

\(60,\!754.80\)
Suppose an ordinary annuity offers an annual interest rate of \(2 \%\), compounded monthly, for 30 years. What should the monthly payment be to have \(\$100,\!000\) at the end of the term?

For \(\$100,\!000\), the monthly payment is \(\approx \$202.95\).
In this exercise, use Theorem geoseries to represent \(f(x) = \frac {1}{x^2+4}\) as a series.
Show that \(f(x) = \frac {\frac {1}{4}}{1 - \left ( -\frac {x^2}{4} \right )}\).
Use the formula in Theorem geoseries: \(\frac {a}{1-r} = {\sum _{k=1}^{\infty } ar^{k-1}}\) to write \(f(x)\) as an infinite series.

\(f(x) = \frac {\frac {1}{4}}{1 - \left ( -\frac {x^2}{4} \right )} = \ds {\sum _{k=1}^{\infty } \frac {1}{4} \,\left ( -\frac {x^2}{4} \right )^{k-1}} = \ds {\sum _{k=1}^{\infty } \frac {(-1)^{k-1} \, x^{2k-2}}{4^{k}}} \)
Graph \(y = f(x)\) along with some partial sums of the series. What do you notice?

No matter how many terms are added, the graph of the series seems to only account for a portion of the graph of \(y = f(x)\). This is due to the fact that geometric series converge only when the ratio \(|r| < 1\). In this case, \(r = -\frac {x^2}{4}\) so \(|r| < 1\) corresponds to the interval \((-2,2)\).
Using Example rightsumex as a guide, find and simplify formula for the right endpoint sum, \(RS_{n}\), for each of the functions below on the specified interval. Find \(\ds {\lim _{n \rightarrow \infty } RS_{n}}\) to find the area between the graph of \(f\) and the \(x\)-axis.
\(f(x) = 4-x\) over the interval \([0,4]\).

\(RS_{n} = 8 - \frac {8}{n}\); Area is 8 \(\text {units}^2\)
\(f(x) = 3x^2\) over the interval \([1,3]\).

\(RS_{n} = 26 + \frac {24}{n} + \frac {4}{n^2}\); Area is 26 \(\text {units}^2\)
\(f(x) = 12-x-x^2\) over the interval \([0,3]\).

\(RS_{n} = \frac {45}{2} - \frac {18}{n} - \frac {9}{2n^2}\); Area is \(\frac {45}{2}\) \(\text {units}^2\)
Prove the properties listed in Theorem sigmaprops.
Show that the formula for the future value of an annuity due is
\[A = P(1 + i)\left [\frac {(1 + i)^{nt} - 1}{i}\right ]\]
Discuss with your classmates what goes wrong when trying to find the following sums.
\( { \sum _{k=1}^{\infty } 2^{k-1}}\)
\( { \sum _{k=1}^{\infty } (1.0001)^{k-1}}\)
\( { \sum _{k=1}^{\infty } (-1)^{k-1}}\)
In this exercise, we walk through the proof of Cauchy’s Bound, Theorem CauchysBound in Section RealZeros.

Let \(f(x) = a_{n} x^{n} + a_n-1x^n-1 + \ldots + a_1 x + a_0\) be a polynomial of degree \(n\) and let \(Z\) be the largest zero of \(f\) in absolute value and let \(M\) be the largest of the numbers: \(\frac {|a_{0}|}{|a_{n}|}\), \(\frac {|a_{1}|}{|a_{n}|}\), …, \(\frac {|a_{n-1}|}{|a_{n}|}\).

Since \(P(Z) = 0\), solve for \(Z^{n}\): \(Z^{n} = \frac {a_n-1}{a_{n}} \, Z^n-1 + \ldots + \frac {a_1}{a_{n}} \, Z + \frac {a_0}{a_{n}}\).
If \(-1 \leq Z \leq 1\), then Cauchy’s Bound is immediately satisfied since \(Z\) would automatically lie in the interval \(\left [-(M+1), M+1\right ]\). So we assume \(|Z|>1\). Under the assumption \(|Z|>1\). explain why
\[\begin{array}{rcl} |Z|^{n} & = & \left | \frac {a_n-1}{a_{n}} \, Z^n-1 + \ldots + \frac {a_1}{a_{n}} \, Z + \frac {a_0}{a_{n}} \right | \\[10pt] & \leq & \frac {|a_{n-1}|}{|a_{n}|} \, |Z|^{n-1} + \ldots + \frac {|a_{1}|}{|a_{n}|} \, |Z| + \frac {|a_{0}|}{|a_{n}|} \\ \end{array} \]
Use the definition of \(M\) along with the Geometric Sum Formula, Equation arithgeosum to show:
\[ |Z|^{n} \leq M \left ( |Z|^{n-1} + \ldots + |Z| + 1\right ) = M \, \frac {1 - |Z|^n}{1 - |Z|} = M \, \frac { |Z|^n - 1}{|Z| - 1} \]
Now use the fact that \(|Z| > 1\) to rearrange the above inequality to get:
\[ |Z| - 1 \leq M \, \frac {|Z|^{n} - 1}{|Z|^n} = M \left ( 1 - \frac {1}{|Z|^n} \right ) \]
Use the fact that \(1 - \frac {1}{|Z|^n} < 1\) to get:
\[ |Z| - 1 \leq M \left ( 1 - \frac {1}{|Z|^n} \right ) < M (1) = M\]
From \(|Z| - 1 < M\), we get \(|Z| < M+1\). Hence, \(Z\) lies in the interval \(\left [-(M+1), M+1\right ]\).