2.4 Sequences and Summations

Ex: 10, 11

  • Sequence - a1,a2⋯ana_1, a_2 \cdots a_n
  • Summation - ∑i=1nai=a1+a2+⋯+an\sum^{n}_{i=1} a_i = a_1 + a_2 + \cdots + a_n

Homework

p189: 16cef, 19, 22

16. Find the solution to each of these recurrence relations with the given initial conditions. Use an iterative approach such as that used in Example 10.
c. an=an−1−n,a0=4a_n = a_{n-1} - n, a_0 = 4

Solution

an=[an−2−(n−1)]−n=an−2−2n+1=[an−3−(n−2)]−2n+1=an−3−3n+1+2=⋯=a0−n⋅n+1+2+⋯+n−1=a0−n2+(1+n−1)(n−1)2=4−n2+n2\begin{aligned} a_n &= [a_{n-2}-(n-1)] - n\\ &= a_{n-2}-2n+1\\ &= [a_{n-3}-(n-2)]-2n+1\\ &= a_{n-3}-3n + 1+ 2\\ &= \cdots\\ &= a_0- n\cdot n + 1 + 2 + \cdots + n-1\\ &= a_0-n^2 + \frac{(1+n-1)(n-1)}{2}\\ &= 4-\frac{n^2+n}{2} \end{aligned}

e. an=(n+1)an−1,a0=2a_n = (n+1)a_{n-1}, a_0 = 2

Solution

an=(n+1)n⋅an−2=(n+1)n(n−1)⋅an−3=⋯=(n+1)n(n−1)⋯2×a0=2(n+1)!\begin{aligned} a_n &= (n+1)n\cdot a_{n-2}\\ &= (n+1)n(n-1) \cdot a_{n-3}\\ &= \cdots\\ &= (n+1)n(n-1)\cdots 2 \times a_0\\ &= 2(n+1)! \end{aligned}

f. an=2n(n+1)an−1,a0=3a_n = 2n(n+1)a_{n-1}, a_0 = 3

Solution

an=2n(n+1)⋅2(n−1)n⋅an−2=22⋅(n+1)n⋅n(n−1)⋅an−2=22⋅(n+1)n⋅n(n−1)⋅2(n−2)(n−1)⋅an−3=23⋅(n+1)n(n−1)⋅n(n−1)(n−2)⋅an−3=⋯=2n⋅[(n+1)n(n−1)⋯2]⋅[n(n−1)(n−2)⋯1]⋯a0=2n⋅(n+1)!⋅n!⋅3\begin{aligned} a_n &= 2n(n+1) \cdot 2(n-1)n \cdot a_{n-2}\\ &= 2^2 \cdot (n+1)n \cdot n(n-1) \cdot a_{n-2}\\ &= 2^2 \cdot (n+1)n \cdot n(n-1) \cdot 2(n-2)(n-1) \cdot a_{n-3}\\ &= 2^3 \cdot (n+1)n(n-1) \cdot n(n-1)(n-2) \cdot a_{n-3}\\ &= \cdots\\ &= 2^n \cdot [(n+1)n(n-1)\cdots 2] \cdot [n(n-1)(n-2)\cdots 1] \cdots a_0\\ &= 2^n \cdot (n+1)! \cdot n! \cdot 3 \end{aligned}

19. Suppose that the number of bacteria in a colony triples every hour.
a) Set up a recurrence relation for the number of bacteria after n hours have elapsed.
b) If 100 bacteria are used to begin a new colony, how many bacteria will be in the colony in 10 hours?

Solution

Let the initial number of bacteria be a0a_0, and the number of bacteria after nn hours be ana_n

an=3an−1=32⋅an−2=33⋅an−3=⋯=3n⋅a0a0=100⇒a10=310⋅100\begin{aligned} a_n &= 3a_{n-1}\\ &= 3^2 \cdot a_{n-2}\\ &= 3^3 \cdot a_{n-3}\\ &=\cdots\\ &= 3^n \cdot a_0\\ a_0&=100 \To a_{10} = 3^{10} \cdot 100 \end{aligned}

22. An employee joined a company in 2009 with a starting salary of $50,000\$50,000. Every year this employee receives a raise of $1000\$1000 plus 5%5\% of the salary of the previous year.
a) Set up a recurrence relation for the salary of this employee n years after 2009.
b) What will the salary of this employee be in 2017?
c) Find an explicit formula for the salary of this employee nn years after 2009.

Solution

Let the salary in 2009 be a0=50000a_0=50000, and the salary after nn years be ana_n.

an=an−1⋅1.05+1000=[an−2⋅1.05+1000]⋅1.05+1000=an−2⋅1.052+1000⋅1.051+1000=[an−3⋅1.05+1000]⋅1.052+1000⋅1.051+1000=an−3⋅1.053+1000⋅1.052+1000⋅1.051+1000=⋯=a0⋅1.05n+1000⋅1.05n−1+⋯+1000⋅1.052+1000⋅1.051+1000=a0⋅1.05n+1000(1.05n−1+⋯+1.050)=a0⋅1.05n+1000⋅1.05n−11.05−1=50000⋅1.05n+20000⋅1.05n−200000=70000⋅1.05n−20000\begin{aligned} a_n &= a_{n-1} \cdot 1.05 + 1000\\ &= [a_{n-2} \cdot 1.05 + 1000] \cdot 1.05 + 1000\\ &= a_{n-2} \cdot 1.05^2 + 1000 \cdot 1.05^1 + 1000\\ &= [a_{n-3} \cdot 1.05 + 1000] \cdot 1.05^2 + 1000 \cdot 1.05^1 + 1000\\ &= a_{n-3} \cdot 1.05^3 + 1000 \cdot 1.05^2 + 1000 \cdot 1.05^1 + 1000\\ &= \cdots \\ &= a_0 \cdot 1.05^n + 1000 \cdot 1.05^{n-1} + \cdots + 1000 \cdot 1.05^2 + 1000 \cdot 1.05^1 + 1000\\ &= a_0 \cdot 1.05^n + 1000(1.05^{n-1} + \cdots + 1.05^0)\\ &= a_0 \cdot 1.05^n + 1000 \cdot \frac{1.05^n-1}{1.05-1}\\ &= 50000 \cdot 1.05^n + 20000 \cdot 1.05^n - 200000\\ &= 70000 \cdot 1.05^n-20000\\ \end{aligned}

When the year is 2017, n=8n=8, a8=70000⋅1.05n−20000≈83421.88a_8= 70000 \cdot 1.05^n-20000 \approx 83421.88.