Download 9.2 Summation Notation and more Slides English in PDF only on Docsity!
9.2 Summation Notation 661
9.2 Summation Notation
In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with a definition, which, while intimidating, is meant to make our lives easier.
Definition 9.3. Summation Notation: Given a sequence {an}∞ n=k and numbers m and p satisfying k ≤ m ≤ p, the summation from m to p of the sequence {an} is written
∑^ p
n=m
an = am + am+1 +... + ap
The variable n is called the index of summation. The number m is called the lower limit of summation while the number p is called the upper limit of summation.
In English, Definition 9.3 is simply defining a short-hand notation for adding up the terms of the sequence {an}∞ n=k from am through ap. The symbol Σ is the capital Greek letter sigma and is shorthand for ‘sum’. The lower and upper limits of the summation tells us which term to start with and which term to end with, respectively. For example, using the sequence an = 2n − 1 for n ≥ 1, we can write the sum a 3 + a 4 + a 5 + a 6 as
∑^6
n=
(2n − 1) = (2(3) − 1) + (2(4) − 1) + (2(5) − 1) + (2(6) − 1)
= 5 + 7 + 9 + 11 = 32
The index variable is considered a ‘dummy variable’ in the sense that it may be changed to any letter without affecting the value of the summation. For instance,
∑^6
n=
(2n − 1) =
∑^6
k=
(2k − 1) =
∑^6
j=
(2j − 1)
One place you may encounter summation notation is in mathematical definitions. For example, summation notation allows us to define polynomials as functions of the form
f (x) =
∑^ n
k=
akxk
for real numbers ak, k = 0, 1 ,... n. The reader is invited to compare this with what is given in Definition 3.1. Summation notation is particularly useful when talking about matrix operations. For example, we can write the product of the ith row Ri of a matrix A = [aij ]m×n and the jth column Cj of a matrix B = [bij ]n×r as
Ri · Cj =
∑^ n
k=
aikbkj
662 Sequences and the Binomial Theorem
Again, the reader is encouraged to write out the sum and compare it to Definition 8.9. Our next example gives us practice with this new notation.
Example 9.2.1.
- Find the following sums.
(a)
∑^4
k=
100 k^ (b)
∑^4
n=
n! 2 (c)
∑^5
n=
(−1)n+ n (x^ −^ 1)
n
- Write the following sums using summation notation.
(a) 1 + 3 + 5 +... + 117
(b) 1 − 1 2
+^1
(c) 0.9 + 0.09 + 0.009 +... 0. (^0) ︸ ︷︷ ︸ · · · 0 n − 1 zeros
Solution.
- (a) We substitute k = 1 into the formula (^10013) k and add successive terms until we reach k = 4.
∑^4
k=
100 k^ =^
1001 +^
1002 +^
1003 +^
(b) Proceeding as in (a), we replace every occurrence of n with the values 0 through 4. We recall the factorials, n! as defined in number Example 9.1.1, number 6 and get:
∑^4
n=
n! 2 =^
+^1
+^2 ·^1
+^3 ·^2 ·^1
+^4 ·^3 ·^2 ·^1
(c) We proceed as before, replacing the index n, but not the variable x, with the values 1 through 5 and adding the resulting terms.
664 Sequences and the Binomial Theorem
the upper limit of summation. Since n is used in the limits of the summation, we need to choose a different letter for the index of summation.^3 We choose k and get
n − 1 zeros
∑^ n
k=
10 k
The following theorem presents some general properties of summation notation. While we shall not have much need of these properties in Algebra, they do play a great role in Calculus. Moreover, there is much to be learned by thinking about why the properties hold. We invite the reader to prove these results. To get started, remember, “When in doubt, write it out!”
Theorem 9.1. Properties of Summation Notation: Suppose {an} and {bn} are sequences so that the following sums are defined.
∑^ p
n=m
(an ± bn) =
∑^ p
n=m
an ±
∑^ p
n=m
bn
∑^ p
n=m
c an = c
∑^ p
n=m
an, for any real number c.
∑^ p
n=m
an =
∑^ j
n=m
an +
∑^ p
n=j+
an, for any natural number m ≤ j < j + 1 ≤ p.
∑^ p
n=m
an =
p∑+r
n=m+r
an−r, for any whole number r.
We now turn our attention to the sums involving arithmetic and geometric sequences. Given an arithmetic sequence ak = a + (k − 1)d for k ≥ 1, we let S denote the sum of the first n terms. To derive a formula for S, we write it out in two different ways
S = a + (a + d) +... + (a + (n − 2)d) + (a + (n − 1)d) S = (a + (n − 1)d) + (a + (n − 2)d) +... + (a + d) + a
If we add these two equations and combine the terms which are aligned vertically, we get
2 S = (2a + (n − 1)d) + (2a + (n − 1)d) +... + (2a + (n − 1)d) + (2a + (n − 1)d)
The right hand side of this equation contains n terms, all of which are equal to (2a + (n − 1)d) so we get 2S = n(2a + (n − 1)d). Dividing both sides of this equation by 2, we obtain the formula
(^3) To see why, try writing the summation using ‘n’ as the index.
9.2 Summation Notation 665
S =
n 2 (2a^ + (n^ −^ 1)d)
If we rewrite the quantity 2a + (n − 1)d as a + (a + (n − 1)d) = a 1 + an, we get the formula
S = n
a 1 + an 2
A helpful way to remember this last formula is to recognize that we have expressed the sum as the product of the number of terms n and the average of the first and nth^ terms.
To derive the formula for the geometric sum, we start with a geometric sequence ak = ark−^1 , k ≥ 1, and let S once again denote the sum of the first n terms. Comparing S and rS, we get
S = a + ar + ar^2 +... + arn−^2 + arn−^1 rS = ar + ar^2 +... + arn−^2 + arn−^1 + arn
Subtracting the second equation from the first forces all of the terms except a and arn^ to cancel out and we get S − rS = a − arn. Factoring, we get S(1 − r) = a (1 − rn). Assuming r 6 = 1, we can divide both sides by the quantity (1 − r) to obtain
S = a
1 − rn 1 − r
If we distribute a through the numerator, we get a − arn^ = a 1 − an+1 which yields the formula
S = a^1 1 − −^ a nr+
In the case when r = 1, we get the formula
S = a︸ + a +︷︷... + a︸ n times
= n a
Our results are summarized below.
9.2 Summation Notation 667
payment at the end of the third compounding period to obtain
A 3 = A 2 (1 + i) + P = P (1 + i)
1 + i
(1 + i) + P = P (1 + i)^2
1 + i
(1 + i)^2
During the fourth compounding period, A 3 grows to A 3 (1+i), and when we add the fourth payment, we factor out P (1 + i)^3 to get
A 4 = P (1 + i)^3
1 + i +^
(1 + i)^2 +^
(1 + i)^3
This pattern continues so that at the end of the kth compounding, we get
Ak = P (1 + i)k−^1
1 + i
(1 + i)^2
(1 + i)k−^1
The sum in the parentheses above is the sum of the first k terms of a geometric sequence with a = 1 and r = (^) 1+^1 i. Using Equation 9.2, we get
1 + i
(1 + i)^2
(1 + i)k−^1
(1 + i)k 1 − 1 1 + i
=^
(1 + i)
1 − (1 + i)−k
i
Hence, we get
Ak = P (1 + i)k−^1
(1 + i)
1 − (1 + i)−k
i
P
(1 + i)k^ − 1
i
If we let t be the number of years this investment strategy is followed, then k = nt, and we get the formula for the future value of an ordinary annuity.
Equation 9.3. Future Value of an Ordinary Annuity: Suppose an annuity offers an annual interest rate r compounded n times per year. Let i = rn be the interest rate per compounding period. If a deposit P is made at the end of each compounding period, the amount A in the account after t years is given by
A =
P
(1 + i)nt^ − 1
i
The reader is encouraged to substitute i = rn into Equation 9.3 and simplify. Some familiar equations arise which are cause for pause and meditation. One last note: if the deposit P is made a the beginning of the compounding period instead of at the end, the annuity is called an annuity- due. We leave the derivation of the formula for the future value of an annuity-due as an exercise for the reader.
668 Sequences and the Binomial Theorem
Example 9.2.2. An ordinary annuity offers a 6% annual interest rate, compounded monthly.
- If monthly payments of $50 are made, find the value of the annuity in 30 years.
- How many years will it take for the annuity to grow to $100,000?
Solution.
- We have r = 0.06 and n = 12 so that i = (^) nr = 012.^06 = 0.005. With P = 50 and t = 30,
A =
(1 + 0.005)(12)(30)^ − 1
0. 005 ≈^50225.^75
Our final answer is $50, 225 .75.
- To find how long it will take for the annuity to grow to $100,000, we set A = 100000 and solve for t. We isolate the exponential and take natural logs of both sides of the equation.
(1 + 0.005)^12 t^ − 1
10 = (1.005)^12 t^ − 1 (1.005)^12 t^ = 11 ln
(1.005)^12 t
= ln(11) 12 t ln(1.005) = ln(11) t = (^) 12 ln(1ln(11).005) ≈ 40. 06
This means that it takes just over 40 years for the investment to grow to $100,000. Comparing this with our answer to part 1, we see that in just 10 additional years, the value of the annuity nearly doubles. This is a lesson worth remembering.
We close this section with a peek into Calculus by considering infinite sums, called series. Consider the number 0.9. We can write this number as
0 .9 = 0. 9999 ... = 0.9 + 0.09 + 0.009 + 0.0009 +...
From Example 9.2.1, we know we can write the sum of the first n of these terms as
n nines
n − 1 zeros
∑^ n
k=
10 k
Using Equation 9.2, we have
