Question

ap precalculus c10 review worksheet name: period group show all work using algebraic methods 1. find the sum algebraically \\(\sum\limits_{n = 3}^{11}\left(\frac{3}{5}(5)^{n - 4}\
ight)\\) 2. find the sum algebraically \\(\sum\limits_{n = - 4}^{9}(2n + 3)\\) 4. find the sum: \\(24 + 36 + 54 + \dots + 121.5\\) 5. find the sum of the first 23 terms of \\(31 + 18 + 5 + \dots\\) 6. express using sigma notation. \\(0.32 + 0.89 + 1.46 + \dots + 8.3\\) 7. find \\(n\\) if \\(s_{n}=872\\) for \\(\sum\limits_{n = 1}^{\infty}(7n - 5)\\) 8. find \\(n\\) if \\(a_{n}=26.2144\\) given \\(125 + 100 + 80 + \dots\\)

Explanation:

Step1: Rewrite the summand

$\frac{3}{5}(5)^{n-4} = 3 \cdot 5^{n-5}$

Step2: Adjust index to start at 0

Let $k = n-3$, so $n=k+3$, $k=0$ to $8$.
$\sum_{k=0}^{8} 3 \cdot 5^{(k+3)-5} = \sum_{k=0}^{8} 3 \cdot 5^{k-2} = \frac{3}{25}\sum_{k=0}^{8}5^k$

Step3: Use finite geometric sum formula

Sum of $\sum_{k=0}^{n}r^k = \frac{r^{n+1}-1}{r-1}$, $r=5$, $n=8$
$\frac{3}{25} \cdot \frac{5^{9}-1}{5-1}$

Step4: Calculate the value

$5^9=1953125$, so $\frac{3}{25} \cdot \frac{1953125-1}{4} = \frac{3}{25} \cdot \frac{1953124}{4} = \frac{3 \cdot 488281}{25} = \frac{1464843}{25} = 58593.72$

Step1: Expand the summation

$\sum_{n=-4}^{9}(2n+3) = 2\sum_{n=-4}^{9}n + \sum_{n=-4}^{9}3$

Step2: Count number of terms

Number of terms: $9 - (-4) +1=14$

Step3: Calculate sum of n values

$\sum_{n=-4}^{9}n = \sum_{k=1}^{14}(k-5) = \sum_{k=1}^{14}k - 5\cdot14 = \frac{14\cdot15}{2}-70=105-70=35$

Step4: Calculate sum of constants

$\sum_{n=-4}^{9}3 = 3\cdot14=42$

Step5: Combine results

$2\cdot35 + 42=70+42=112$

Step1: Identify geometric sequence

$a_1=24$, common ratio $r=\frac{36}{24}=1.5$

Step2: Find number of terms

Solve $24(1.5)^{n-1}=121.5$
$(1.5)^{n-1}=\frac{121.5}{24}=5.0625$
$1.5^5=7.59375$, $1.5^4=5.0625$, so $n-1=4$, $n=5$

Step3: Use geometric sum formula

$S_n=a_1\frac{r^n-1}{r-1}=24\cdot\frac{1.5^5-1}{1.5-1}$

Step4: Compute the value

$1.5^5=7.59375$, so $24\cdot\frac{7.59375-1}{0.5}=24\cdot\frac{6.59375}{0.5}=24\cdot13.1875=316.5$

Step1: Identify arithmetic sequence

$a_1=31$, common difference $d=18-31=-13$

Step2: Use arithmetic sum formula

$S_n=\frac{n}{2}[2a_1+(n-1)d]$, $n=23$

Step3: Substitute values

$S_{23}=\frac{23}{2}[2\cdot31+(23-1)(-13)]$

Step4: Calculate the result

$\frac{23}{2}[62 + 22(-13)]=\frac{23}{2}[62-286]=\frac{23}{2}(-224)=23(-112)=-2576$

Answer:

$58593.72$

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