Math Problem Statement

solve 12-15

Solve problems 12 to 15 involving summation notation, arithmetic sequences, geometric series, and exponential growth.

Solution

Let's solve each question step by step:


12. Find the sum.

k=16(k23)\sum_{k=1}^{6} (k^2 - 3)

This involves evaluating each term and summing them up.

Steps:

  • Calculate k23k^2 - 3 for k=1,2,3,4,5,6k = 1, 2, 3, 4, 5, 6:
    • k=1:123=13=2k = 1: 1^2 - 3 = 1 - 3 = -2
    • k=2:223=43=1k = 2: 2^2 - 3 = 4 - 3 = 1
    • k=3:323=93=6k = 3: 3^2 - 3 = 9 - 3 = 6
    • k=4:423=163=13k = 4: 4^2 - 3 = 16 - 3 = 13
    • k=5:523=253=22k = 5: 5^2 - 3 = 25 - 3 = 22
    • k=6:623=363=33k = 6: 6^2 - 3 = 36 - 3 = 33

Sum the values: 2+1+6+13+22+33=73-2 + 1 + 6 + 13 + 22 + 33 = 73

Answer: 73\boxed{73}


13. Find the sum of an arithmetic sequence.

k=05(1+2.3k)\sum_{k=0}^{5} (1 + 2.3k)

Steps:

  1. Calculate the first few terms:

    • k=0:1+2.3(0)=1k = 0: 1 + 2.3(0) = 1
    • k=1:1+2.3(1)=3.3k = 1: 1 + 2.3(1) = 3.3
    • k=2:1+2.3(2)=5.6k = 2: 1 + 2.3(2) = 5.6
    • k=3:1+2.3(3)=7.9k = 3: 1 + 2.3(3) = 7.9
    • k=4:1+2.3(4)=10.2k = 4: 1 + 2.3(4) = 10.2
    • k=5:1+2.3(5)=12.5k = 5: 1 + 2.3(5) = 12.5
  2. Sum the terms: 1+3.3+5.6+7.9+10.2+12.5=40.51 + 3.3 + 5.6 + 7.9 + 10.2 + 12.5 = 40.5

Answer: 40.5\boxed{40.5}


14. Find the sum of the infinite geometric series.

1+13+19+127+1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots

This is an infinite geometric series with:

  • First term (aa) = 11,
  • Common ratio (rr) = 13\frac{1}{3}.

The formula for the sum of an infinite geometric series is: S=a1rS = \frac{a}{1 - r}

Substitute the values: S=1113=123=32S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}

Answer: 32 or 1.5\boxed{\frac{3}{2} \text{ or } 1.5}


15. Population growth problem.

Pn=45,000(1.05)nP_n = 45,000 (1.05)^n

(a) Find the first five terms of the sequence.

Substitute n=0,1,2,3,4n = 0, 1, 2, 3, 4:

  • P0=45,000(1.05)0=45,000P_0 = 45,000 (1.05)^0 = 45,000
  • P1=45,000(1.05)1=45,000×1.05=47,250P_1 = 45,000 (1.05)^1 = 45,000 \times 1.05 = 47,250
  • P2=45,000(1.05)2=45,000×1.1025=49,612.549,613P_2 = 45,000 (1.05)^2 = 45,000 \times 1.1025 = 49,612.5 \approx 49,613
  • P3=45,000(1.05)3=45,000×1.157625=52,093.12552,093P_3 = 45,000 (1.05)^3 = 45,000 \times 1.157625 = 52,093.125 \approx 52,093
  • P4=45,000(1.05)4=45,000×1.21550625=54,697.7812554,698P_4 = 45,000 (1.05)^4 = 45,000 \times 1.21550625 = 54,697.78125 \approx 54,698

Answers: P1=47,250, P2=49,613, P3=52,093, P4=54,698P_1 = 47,250,\ P_2 = 49,613,\ P_3 = 52,093,\ P_4 = 54,698

(b) Find the population in 2014.

Since 2014 is 10 years after 2004, n=10n = 10: P10=45,000(1.05)10P_{10} = 45,000 (1.05)^{10} P10=45,000×1.628894626777442=73,300.258P_{10} = 45,000 \times 1.628894626777442 = 73,300.258

Rounded to the nearest whole number: 73,300\boxed{73,300}


Let me know if you'd like further details or have any questions! Here are 5 related questions to explore:

  1. How do you simplify summations using properties?
  2. What happens to a geometric series if the common ratio r1|r| \geq 1?
  3. Can you calculate the growth rate for other exponential models?
  4. How do you find the nn-th term in an arithmetic sequence?
  5. What is the difference between finite and infinite series?

Tip: For population growth, exponential models are useful for estimating future sizes accurately!

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Math Problem Analysis

Mathematical Concepts

Summation Notation
Arithmetic Sequence
Geometric Series
Exponential Growth

Formulas

Sum of the first n terms of an arithmetic sequence: S = (n/2)(a + l)
Sum of an infinite geometric series: S = a / (1 - r)
Exponential growth model: P_n = P_0(1 + r)^n

Theorems

Properties of summation
Formula for the sum of an arithmetic series
Formula for the sum of an infinite geometric series

Suitable Grade Level

Grades 9-12