1.
Find the exact value of the sum for the following series: \(\sum_{k=1}^{7}3(2)^{k-1}\)
Step 1. Evaluate the terms from \(k=1\) to \(k=7\). \[ 3(2)^{k-1} \] Step 2. Substitute each value of \(k\). \[ \begin{aligned} k=1:&\quad 3(2)^{1-1}=3(2)^0=3\\k=2:&\quad 3(2)^{2-1}=3(2)^1=6\\k=3:&\quad 3(2)^{3-1}=3(2)^2=12\\k=4:&\quad 3(2)^3=24\\k=5:&\quad 3(2)^4=48\\k=6:&\quad 3(2)^5=96\\k=7:&\quad 3(2)^6=192\end{aligned} \] Step 3. Add the terms. \[ 3+6+12+24+48+96+192=381 \] Therefore, the exact value of the sum is \(381\). Answer: D
2.
Find the number of terms for the following series: \(\sum_{x=-2}^{5}\frac{1}{4}\left(-\frac{1}{2}\right)^x\)
Step 1. Look at the limits of the sigma notation. \[ \sum_{x=-2}^{5}\frac{1}{4}\left(-\frac{1}{2}\right)^x \] Step 2. The lower limit is \(x=-2\), so the terms start at \(-2\). Step 3. The upper limit is \(5\), so the terms stop at \(5\). Step 4. List the values of \(x\). \[ x=-2,-1,0,1,2,3,4,5 \] Step 5. Count the terms. \[ 8\text{ terms} \] Therefore, there are \(8\) terms in this series. Answer: E
3.
A hard rubber ball is dropped from a roof of a building that is \(4\) meters high. The ball rises to \(40\%\) of the height from which it fell after each bounce. What is the total vertical distance that the ball had traveled by the time it hit the ground for the \(10\)th time?
Step 1. The ball first falls \(4\text{ m}\). Step 2. After each bounce, it rises to \(40\%\) of the previous height, so: \[ r=0.4 \] Step 3. The first bounce height is: \[ a=4(0.4)=1.6 \] Step 4. By the time it hits the ground for the \(10\)th time, use \(n=9\) bounce heights after the initial drop. Step 5. Use the finite geometric sum formula. \[ S_9=\frac{a(1-r^9)}{1-r} \] Step 6. Substitute \(a=1.6\) and \(r=0.4\). \[ S_9=\frac{1.6(1-0.4^9)}{1-0.4} \] Step 7. The total vertical distance is the initial drop plus twice the bounce-height sum. \[ \begin{aligned} \text{Total distance}&=4+2S_9\\&=4+2\left(\frac{1.6(1-0.4^9)}{1-0.4}\right)\\&\approx 9.33\end{aligned} \] Therefore, the total vertical distance is approximately \(9.33\text{ m}\). Answer: A
4.
Determine the common ratio of the following geometric sequence: \(5,-25,125,-625,\cdots\)
Step 1. To find the common ratio, divide a term by the previous term. \[ r=\frac{t_2}{t_1} \] Step 2. Substitute the first two terms. \[ r=\frac{-25}{5}=-5 \] Step 3. Check with the next terms. \[ \frac{125}{-25}=-5,\quad \frac{-625}{125}=-5 \] Therefore, the common ratio is \(-5\). Answer: C
5.
Two years after purchase, the resale value of a car was \(\$10000\). The resale value of the same car three years later was \(\$5000\). If the annual depreciation of the car forms a geometric sequence, what was the original price of the car?
Step 1. Let \(a\) be the original price of the car and \(r\) be the annual depreciation ratio. Step 2. Two years after purchase, the value is \(10000\). \[ ar^2=10000 \] Step 3. Three years later means five years after purchase, the value is \(5000\). \[ ar^5=5000 \] Step 4. Divide the two equations to solve for \(r\). \[ \frac{ar^5}{ar^2}=\frac{5000}{10000} \] \[ r^3=0.5 \] \[ r=\sqrt[3]{0.5} \] Step 5. Use \(ar^2=10000\) to solve for \(a\). \[ a=\frac{10000}{r^2} \] Step 6. Substitute \(r=\sqrt[3]{0.5}\). \[ a=\frac{10000}{\left(\sqrt[3]{0.5}\right)^2}\approx 15874.01 \] Therefore, the original price of the car was approximately \(\$15874.01\). Answer: B
6.
If the sum of an infinite geometric series is \(\frac{100}{9}\) and the common ratio is \(\frac{1}{10}\), determine the first term.
Step 1. Use the infinite geometric series formula. \[ S_{\infty}=\frac{a}{1-r} \] Step 2. Substitute \(S_{\infty}=\frac{100}{9}\) and \(r=\frac{1}{10}\). \[ \frac{100}{9}=\frac{a}{1-\frac{1}{10}} \] Step 3. Simplify the denominator. \[ 1-\frac{1}{10}=\frac{9}{10} \] Step 4. Solve for \(a\). \[ \frac{100}{9}=\frac{a}{\frac{9}{10}} \] \[ a=\frac{100}{9}\cdot\frac{9}{10}=10 \] Therefore, the first term is \(10\). Answer: C
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