Solution Bank Sequences and Series Target B
Answers to the Practice Problems are below in random order.
\(\begin{align}&342\ & \ \ &125250\ & \ \ &a_n=5-(n-1)\\\\
&\ln (\frac{8 \sqrt{z}}{y})\ & \ \ &331\ & \ \ &a_n=11+4(n-1)\\\\
&\text{No, change is multiplicative}\ & \ \ &f^{-1}(x)=e^{\frac{x}{5}}+3\ & \ \ &\log_{2} 9,\ \text{since} 2^3 <9\ \text{and}\ 3^3 > 16\\\\
&a_n=4+3(n-1)\ & \ \ &\text{Yes,}\ d=-\frac{2}{3}\ & \ \ &795\\\\
&-34875\ & \ \ &a_n=9+7(n-1)\ & \ \ &\$2450\\\\
&a_n=n\ & \ \ &a_n=-8+6(n-1)\ & \ \ &\text{Dom:}\ x < 3\ \text{Rng:}\ y \geq 0\\\\
&a_n=34-6(n-1)\ & \ \ &a_n=-35+6(n-1)\ & \ \ &a_n=5\\\\
&32.58\ \text{grams}\ & \ \ &a_n=4+5(n-1)\ & \ \ &561\\\\
&\text{Yes,}\ d=-4\end{align}\)
Answers to the Practice Problems are below in random order.
\(\begin{align}&342\ & \ \ &125250\ & \ \ &a_n=5-(n-1)\\\\
&\ln (\frac{8 \sqrt{z}}{y})\ & \ \ &331\ & \ \ &a_n=11+4(n-1)\\\\
&\text{No, change is multiplicative}\ & \ \ &f^{-1}(x)=e^{\frac{x}{5}}+3\ & \ \ &\log_{2} 9,\ \text{since} 2^3 <9\ \text{and}\ 3^3 > 16\\\\
&a_n=4+3(n-1)\ & \ \ &\text{Yes,}\ d=-\frac{2}{3}\ & \ \ &795\\\\
&-34875\ & \ \ &a_n=9+7(n-1)\ & \ \ &\$2450\\\\
&a_n=n\ & \ \ &a_n=-8+6(n-1)\ & \ \ &\text{Dom:}\ x < 3\ \text{Rng:}\ y \geq 0\\\\
&a_n=34-6(n-1)\ & \ \ &a_n=-35+6(n-1)\ & \ \ &a_n=5\\\\
&32.58\ \text{grams}\ & \ \ &a_n=4+5(n-1)\ & \ \ &561\\\\
&\text{Yes,}\ d=-4\end{align}\)