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: C
2.
A person has \(2\) parents one generation ago and \(4\) grandparents two generations ago and so on. How many ancestors does a person have in the \(9\)th generation?
Step 1. The number of ancestors forms a geometric sequence. \[ 2,4,8,16,\cdots \] Step 2. Identify the first term and common ratio. \[ a=2,\quad r=2 \] Step 3. Use the geometric sequence formula. \[ a_n=ar^{n-1} \] Step 4. Substitute \(n=9\). \[ a_9=2(2)^{9-1} \] Step 5. Simplify. \[ a_9=2(2^8)=2(256)=512 \] Therefore, the person has \(512\) ancestors in the \(9\)th generation. Answer: C
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.
Find the number of terms for the following series: \(\sum_{k=1}^{7}3(2)^{k-1}\)
Step 1. Look at the limits of the sigma notation. \[ \sum_{k=1}^{7}3(2)^{k-1} \] Step 2. The lower limit is \(k=1\), so the terms start at \(1\). Step 3. The upper limit is \(7\), so the terms stop at \(7\). Step 4. The values of \(k\) are: \[ k=1,2,3,4,5,6,7 \] Therefore, there are \(7\) terms in the series. Answer: A
5.
If the sum of an infinite geometric series is \(81\) and the common ratio is \(\frac{2}{3}\), determine the first term.
Step 1. Use the infinite geometric series formula. \[ S_{\infty}=\frac{a}{1-r} \] Step 2. Substitute \(S_{\infty}=81\) and \(r=\frac{2}{3}\). \[ 81=\frac{a}{1-\frac{2}{3}} \] Step 3. Simplify the denominator. \[ 1-\frac{2}{3}=\frac{1}{3} \] Step 4. Solve for \(a\). \[ \begin{aligned} 81&=\frac{a}{\frac{1}{3}}\\81&=3a\\a&=27\end{aligned} \] Therefore, the first term is \(27\). Answer: A
6.
Determine the common ratio of the following geometric sequence: \(2,-6,18,-54,\cdots\)
Step 1. The common ratio of a geometric sequence is found by dividing a term by the previous term. \[ r=\frac{t_n}{t_{n-1}} \] Step 2. Use the first two terms \(2\) and \(-6\). \[ r=\frac{-6}{2}=-3 \] Step 3. Check the next terms. \[ \frac{18}{-6}=-3,\quad \frac{-54}{18}=-3 \] Therefore, the common ratio is \(-3\). Answer: B
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