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Section 9.3 Conjectures of Explicit Formulas for Sums and Products

Although we have a notation for sums and products, we still need to go through a process to evaluate a sum and product. First, we need to calculate each term. Next, we need to add or multiply. In some cases, we can find an explicit formula for a sum or a product, which allows us to evaluate using a formula. In this section, we introduce partial sums and partial products and use them to conjecture explicit formulas.

Subsection 9.3.A Sum Conjectures

We start with sums.

Definition 9.3.1 Partial sums.

Consider a list of numbers \(a_s, a_{s+1}, a_{s+2}, \ldots, a_n\) for some integers \(s\) and \(n\text{.}\)
  1. For the sum
    \begin{equation*} \sum_{k=s}^n a_k =a_s + a_{s+1}+a_{s+2}+\ldots+a_n\text{,} \end{equation*}
    the partial sums are
    \begin{align*} \sum_{k=s}^s a_k \amp =a_s,\\ \sum_{k=s}^{s+1} a_k \amp =a_s+a_{s+1},\\ \sum_{k=s}^{s+2} a_k \amp =a_s+a_{s+1}+a_{s+2}\text{,} \end{align*}
    and so on. For example, for the sum
    \begin{equation*} \sum_{k=1}^n 2 =2 + 2 + 2 + \ldots + 2\text{,} \end{equation*}
    the partial sums are
    \begin{align*} \sum_{k=1}^1 2 \amp =2,\\ \sum_{k=1}^2 2 \amp =2+2=4,\\ \sum_{k=1}^3 2 \amp =2+2+2=6\text{,} \end{align*}
    and so on.
  2. The sequence of partial sums is
    \begin{equation*} a_s, a_s+a_{s+1},a_s+a_{s+1}+a_{s+2}, \ldots\text{.} \end{equation*}
    For example, the sequence of partial sums in our example begins
    \begin{equation*} 2, 4, 6, \ldots\text{.} \end{equation*}
We focus on finite sums, but we mention, for readers who have studied calculus, that these are the same partial sums used to define infinite series.
There is a shorthand notation for calculating partial sums that we introduce in our next example, where we revisit the sum from Definition 9.3.1.

Example 9.3.2 Conjecture explicit form of sum of constant.

Consider the sum
\begin{equation*} \sum_{k=1}^n 2 =2 + 2 + 2 + \ldots + 2\text{.} \end{equation*}
  1. Calculate the partial sums.
    Solution.
    We display the partial sums using a shorthand in Figure 9.3.3. At each \(+\) sign, we draw a line and record the sum from the beginning to that line.
    Figure 9.3.3. Calculating partial sums for \(\displaystyle \sum_{k=1}^n2\text{.}\)
  2. Display the partial sums in a table.
    Solution.
    We are interested in finding an explicit formula sum as a function of \(n\text{,}\) so it is helpful to have the partial sums in a table, as shown in Table 9.3.4.
    Table 9.3.4. Partial sums for \(\displaystyle \sum_{k=1}^n2\)
    \(n\) 1 2 3 4 5 6
    partial sum 2 4 6 8 10 12
    pattern 2(1) 2(2) 2(3) 2(4) 2(5) 2(6)
  3. Use pattern recognition to conjecture an explicit form.
    Solution.
    We notice that each partial sum is twice the index \(n\text{,}\) and add another row to our table to show the pattern. We conjecture \(\displaystyle \sum_{k=1}^n2=2n\text{,}\) which is no surprise since
    \begin{equation*} \sum_{k=1}^n2 = \underbrace{2 +2+ \ldots + 2}_{n \text{ times } }=2n \end{equation*}
Observe that we can use each partial sum to calculate the next partial sum without having to add from the beginning. For example, once we calculate \(\displaystyle \sum_{k=1}^32 = 2+2+2 = 6\text{,}\) we can calculate
\begin{equation*} \displaystyle \sum_{k=1}^42 = 2+2+2+2 = (2+2+2)+2=6+2=8\text{.} \end{equation*}
That is, the new partial sum is equal to the previous partial sum plus the new term. We state this result in general.

Remark 9.3.5 Recursive definition of partial sums.

Consider a list of numbers \(a_s, a_{s+1}, a_{s+2}, \ldots, a_n\) for some integers \(s\) and \(n\text{.}\) Then
\begin{align*} \sum_{k=s}^n a_k \amp = a_s + a_{s+1} + \ldots + a_{n-1} + a_n\\ \amp = (a_s + a_{s+1} + \ldots + a_{n-1}) + a_n\\ \amp = \left(\sum_{k=s}^{n-1} a_k\right) + a_n \end{align*}
We can remember this recursion as
\begin{equation*} \underbrace{\sum_{k=s}^n a_k}_{\text{ new sum } } = \underbrace{\sum_{k=s}^{n-1} a_k}_{\text{ previous sum } } + \underbrace{a_n}_{\text{ new term } } \end{equation*}
Let’s look at a couple more examples of calculating partial sums and using pattern recognition to conjecture an explicit formula.

Example 9.3.6 Conjecture explicit formula for sum.

Calculate the first few partial sums, display your findings in a table, and use pattern recognition to conjecture an explicit formula for the sum.
  1. \(\displaystyle \sum_{k=1}^n(4k-1)\) Hint: Factor.
    Solution.
    We start by writing out the first six terms and their partial sums, as shown in Figure 9.3.7.
    Figure 9.3.7. Calculating partial sums for \(\displaystyle \sum_{k=1}^n(4k-1)\text{.}\)
    For example, \(3+7 = 10\text{.}\) Next, using the previous sum (as in Remark 9.3.5) we get \(3+7+11 = (10) + 11 = 21\text{.}\) Then, \(21 + 15 = 36\text{,}\) and so on.
    Next, we make a table showing the index \(n\) and the partial sums, as shown in Table 9.3.8.
    Table 9.3.8. Partial sums for \(\displaystyle \sum_{k=1}^n(4k-1)\)
    \(n\) 1 2 3 4 5 6
    partial sum 3 10 21 36 55 78
    pattern 1(3) 2(5) 3(7) 4(9) 5(11) 6(13)
    To see the pattern, we follow the hint and factor each partial sum. We show this pattern in the third row of the table. Notice that one factor is \(n\) itself. The other factor is odd. Based on this pattern, we conjecture that
    \begin{equation*} \displaystyle \sum_{k=1}^n(4k-1)=n(2n+1) \end{equation*}
  2. \(\displaystyle \sum_{k=0}^n9 \cdot 10^k\)
    Solution.
    We start by writing out the first four terms and their partial sums, as shown in Figure 9.3.9.
    Figure 9.3.9. Calculating partial sums for \(\displaystyle \sum_{k=0}^n9 \cdot 10^k\text{.}\)
    For example, using Remark 9.3.5, we calculate \(9+90=99\text{,}\) then \(99 + 900 = 999\text{,}\) and so on.
    Next, we make a table showing the index \(n\) and the partial sums, as shown in Table 9.3.10.
    Table 9.3.10. Partial sums for \(\displaystyle \sum_{k=0}^n9 \cdot 10^k\)
    \(n\) 1 2 3 4
    partial sum 9 99 999 9999
    pattern \(10^1-1\) \(10^2-1\) \(10^3-1\) \(10^4-1\)
    We notice that the partial sums are related to powers of 10 and, in fact, they are each one less than the next higher power of 10. We show this pattern in the third row of the table. Based on this pattern, we conjecture that
    \begin{equation*} \displaystyle \sum_{k=0}^n9 \cdot 10^k=10^{n+1}-1\text{.} \end{equation*}
It is your turn to conjecture explicit forms of a sum.

Activity 9.3.A Sum conjectures.

For each problem, calculate the first few partial sums and use pattern recognition to complete a conjecture about an explicit formula for the sum.
  1. \(\displaystyle \sum_{k=0}^n2^k = 1+2+4+\ldots +2^n\) Hint: The answer involves powers of two.
  2. \(\displaystyle \sum_{k=1}^n(2k)=2+4+6+ \ldots + 2n\) Hint: Factor.
  3. \(\displaystyle \sum_{k=1}^n\frac{2}{k(k+1)}= 1 + \frac{1}{3} + \frac{1}{6} + \ldots+\frac{2}{n(n+1)}\) Use computational technology to find the partial sums as fractions. (In DESMOS, click the fraction icon to ask for the answer as a fraction.)
Let’s look at Activity 9.3.A (c).

Example 9.3.12 Explicit form of sum of reciprocals of handshake numbers.

Calculate the first few partial sums and use pattern recognition to complete a conjecture about an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{k=1}^n\frac{2}{k(k+1)}= 1 + \frac{1}{3} + \frac{1}{6} + \ldots+\frac{2}{n(n+1)} \end{equation*}
Solution.
The first five partial sums are
\begin{equation*} 1, \frac{4}{3}, \frac{3}{2}, \frac{8}{5}, \frac{5}{3}, \frac{12}{7}\text{.} \end{equation*}
Notice the denominators are of the form
\begin{equation*} \underline{\hspace{.2in}}, 3, \underline{\hspace{.2in}}, 5, \underline{\hspace{.2in}}, 7 \end{equation*}
which suggests that we should rewrite the fractions over the denominators
\begin{equation*} 2, 3, 4, 5, 6, 7 \end{equation*}
to get
\begin{equation*} \frac{2}{2}, \frac{4}{3}, \frac{6}{4}, \frac{8}{5}, \frac{10}{6}, \frac{12}{7}\text{.} \end{equation*}
Next, we make a table showing the index \(n\) and the partial sums, as shown in Table 9.3.13.
Table 9.3.13. Partial sums for \(\displaystyle \sum_{k=1}^n\frac{2}{k(k+1)}\)
\(n\) 1 2 3 4 5 6
partial sum 1 \(\displaystyle \frac{4}{3}\) \(\displaystyle\frac{3}{2}\) \(\displaystyle\frac{8}{5}\) \(\displaystyle\frac{5}{3}\) \(\displaystyle\frac{12}{7}\)
pattern \(\displaystyle\frac{2}{2}\) \(\displaystyle\frac{4}{3}\) \(\displaystyle\frac{6}{4}\) \(\displaystyle\frac{8}{5}\) \(\displaystyle\frac{10}{6}\) \(\displaystyle\frac{12}{7}\)
Based on this pattern, we conjecture that
\begin{equation*} \sum_{k=1}^n\frac{2}{k(k+1)}= \frac{2n}{n+1} \end{equation*}

Subsection 9.3.B Product Conjectures

Next, we work with products.

Definition 9.3.14 Partial products.

Consider a list of numbers \(a_s, a_{s+1}, a_{s+2}, \ldots, a_n\) for some integers \(s\) and \(n\text{.}\)
  1. For the product
    \begin{equation*} \prod_{k=s}^n a_k =a_s \times a_{s+1} \times a_{s+2} \times \ldots \times a_n\text{,} \end{equation*}
    the partial products are
    \begin{align*} \prod_{k=s}^s a_k \amp =a_s,\\ \prod_{k=s}^{s+1} a_k \amp =a_s \times a_{s+1},\\ \prod_{k=s}^{s+2} a_k \amp =a_s \times a_{s+1} \times a_{s+2}\text{,} \end{align*}
    and so on. For example, for the product
    \begin{equation*} \prod_{k=1}^n 2 = 2 \times 2 \times \cdots \times 2\text{,} \end{equation*}
    the partial products are
    \begin{align*} \prod_{k=1}^1 2 \amp =2,\\ \prod_{k=1}^2 2 \amp =2 \times 2=4,\\ \prod_{k=1}^3 2 \amp =2 \times 2 \times 2=8\text{,} \end{align*}
    and so on.
  2. The sequence of partial products is
    \begin{equation*} a_s, a_sa_{s+1},a_sa_{s+1}a_{s+2}, \ldots\text{.} \end{equation*}
    For example, the sequence of partial products in our example begins
    \begin{equation*} 2, 4, 8, \ldots\text{.} \end{equation*}
There is a shorthand notation for calculating partial products that we introduce in our next example, where we revisit the product from Definition 9.3.14.

Example 9.3.15 Conjecture explicit form of product of constant.

Consider the product
\begin{equation*} \prod_{k=1}^n 2 =2 \times 2 \times 2 \times \ldots \times 2\text{.} \end{equation*}
  1. Calculate the partial products.
    Solution.
    We display the partial products using a shorthand in Figure 9.3.16. At each × sign, we draw a line and record the product from the beginning to that line.
    Figure 9.3.16. Calculating partial products for \(\displaystyle \prod_{k=1}^n2\text{.}\)
  2. Display the partial products in a table.
    Solution.
    We are interested in finding an explicit formula for the product as a function of \(n\text{,}\) so it is helpful to have the partial products in a table, as shown in Table 9.3.17.
    Table 9.3.17. Partial products for \(\displaystyle \prod_{k=1}^n2\)
    \(n\) 1 2 3 4 5 6
    partial product 2 4 8 16 32 64
    pattern \(2^1\) \(2^2\) \(2^3\) \(2^4\) \(2^5\) \(2^6\)
  3. Use pattern recognition to conjecture an explicit form.
    Solution.
    We notice that each partial product is two to the index \(n\text{,}\) and add another row to our table to show the pattern. We conjecture \(\displaystyle \prod_{k=1}^n2=2^n\text{,}\) which is no surprise since
    \begin{equation*} \prod_{k=1}^n2 = \underbrace{2 \times 2 \times \ldots \times 2}_{n \text{ times } }=2^n \end{equation*}
Observe that we can use each partial product to calculate the next partial product without having to multiply from the beginning. For example, once we calculate \(\displaystyle \prod_{k=1}^32 = 2 \times 2 \times 2 = 8\text{,}\) we can calculate
\begin{equation*} \displaystyle \prod_{k=1}^42 = 2 \times 2 \times 2 \times 2 = (2 \times 2 \times 2)\times 2=8 \times 2=16\text{.} \end{equation*}
That is, the new partial product is equal to the previous partial product times the new term. We state this result in general.

Remark 9.3.18 Recursive definition of partial product.

Consider a list of numbers \(a_s, a_{s+1}, a_{s+2}, \ldots, a_n\) for some integers \(s\) and \(n\text{.}\) Then
\begin{align*} \prod_{k=s}^n a_k \amp = a_s \times a_{s+1} \times \ldots \times a_{n-1} \times a_n\\ \amp = (a_s \times a_{s+1}\times \ldots \times a_{n-1}) \times a_n\\ \amp = \left(\prod_{k=s}^{n-1} a_k\right) \times a_n \end{align*}
We can remember this recursion as
\begin{equation*} \underbrace{\prod_{k=s}^n a_k}_{\text{ new product } } = \underbrace{\prod_{k=s}^{n-1} a_k}_{\text{ previous product } } \times \underbrace{a_n}_{\text{ new term } } \end{equation*}
Let’s look at another example of calculating partial products and using pattern recognition to conjecture an explicit formula.

Example 9.3.19 Conjecture explicit formula for product.

Calculate the first few partial products, display your findings in a table, and use pattern recognition to conjecture an explicit formula for the product: \(\displaystyle \prod_{k=1}^nk\text{.}\)
Solution.
We start by writing out the first six terms and their partial products, as shown in Figure 9.3.20.
Figure 9.3.20. Calculating partial products for \(\displaystyle \prod_{k=1}^nk\text{.}\)
For example, \(1 \times 2 = 2\text{.}\) Next, using the previous product (as in Remark 9.3.18) we get \(1 \times 2 \times 3 = (2) \times 3 = 6\text{.}\) Then, \(6 \times 4 = 24\text{,}\) and so on.
Next, we make a table showing the index \(n\) and the partial products, as shown in Table 9.3.21. We recognize these numbers as the factorials (or we realize that we are multiplying consecutive integers). We show this pattern in the third row of the table.
Table 9.3.21. Partial products for \(\displaystyle \prod_{k=1}^nk\)
\(n\) 1 2 3 4 5 6
partial product 1 2 6 24 120 720
pattern \(1!\) \(2!\) \(3!\) \(4!\) \(5!\) \(6!\)
Based on this pattern, we conjecture that \(\displaystyle \prod_{k=1}^nk=n!\text{,}\) which is no surprise because \(1 \times 2 \times 3 \times \cdots \times n = n!\text{.}\)
It is your turn to conjecture explicit forms of a product.

Activity 9.3.B Product conjectures.

For each problem, calculate the first few partial products and use pattern recognition to complete a conjecture about an explicit formula for the product.
  1. \(\displaystyle \prod_{j=1}^n \frac{j+1}{j} = \frac{2}{1} \times \frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{n+1}{n}\text{.}\)
  2. \(\displaystyle \prod_{k=1}^n(2k) = 2(1) \times 2(2) \times 2(3) \times \cdots \times 2(n)\text{.}\) Hint: Keep the partial products factored. Group together the twos to write one factor as a power of two.

Subsection 9.3.C Detour: Fibonacci Sums

There are some interesting identities involving sums of Fibonacci numbers. Try an example.

Activity 9.3.C Fibonacci sums conjectures.

Recall that the Fibonacci numbers are defined by \(F_0=0\text{,}\) \(F_1=1\text{,}\) and \(F_n = F_{n-1}+F_{n-2}\) for \(n \ge 2\text{.}\) The first few Fibonacci numbers are listed in Table 9.3.24.
Table 9.3.24. The first fifteen Fibonacci numbers
\(F_0=0\) \(F_1=1\) \(F_2=1\) \(F_3=2\) \(F_4=3\)
\(F_5=5\) \(F_6=8\) \(F_7=13\) \(F_8=21\) \(F_9=34\)
\(F_{10}=55\) \(F_{11}=89\) \(F_{12} = 144\) \(F_{13} = 233\) \(F_{14} = 377\)
  1. Calculate the first six partial sums and use pattern recognition to conjecture an explicit formula for the sum
    \begin{align*} \sum_{k=1}^nF_{2k} \amp = F_2+F_4+F_6+F_8+F_{10}+F_{12}+\ldots +F_{2n}\\ \amp = 1+3+8+21+55+144+ \ldots +F_{2n} \end{align*}
    Hint: Add one to each partial sum to see the pattern.
  2. Calculate the first six partial sums and use pattern recognition to conjecture an explicit formula for the sum
    \begin{align*} \sum_{k=1}^n(-1)^{k+1}F_k \amp = F_1-F_2+F_3-F_4+F_5-F_6+ \ldots +(-1)^{n+1}F_n\\ \amp = 1-1+2-3+5-8+ \ldots +F_{2n} \end{align*}
    Hint: Subtract one from each partial sum to see the pattern.
Let’s take a closer look at the sum from Activity 9.3.C (a).

Example 9.3.25 Fibonacci sum conjecture.

Calculate the first six partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{align*} \sum_{k=1}^nF_{2k} \amp = F_2+F_4+F_6+F_8+F_{10}+F_{12}+ \ldots +F_{2n}\\ \amp = 1+3+8+21+55+144 \ldots +F_{2n} \end{align*}
The first fifteen Fibonacci numbers appear in Table 9.3.24.
Solution.
We start by writing out the first six terms and their partial sums, as shown in Figure 9.3.26.
Figure 9.3.26. Calculating partial sums for \(\displaystyle \sum_{k=1}^nF_{2k}\text{.}\)
For example, \(1+3 = 4\text{.}\) Next, using the previous sum (as in Remark 9.3.5) we get \(1+3+8 = (4) + 8 = 12\text{.}\) Then, \(12 + 21 = 33\text{,}\) and so on.
Next, we make a table showing the index \(n\) and the partial sums, as shown in Table 9.3.27.
Table 9.3.27. Partial sums for \(\displaystyle \sum_{k=1}^nF_{2k}\)
\(n\) 1 2 3 4 5 6
partial sum 1 4 12 33 88 232
pattern \(2-1\) \(5-1\) \(13-1\) \(34-1\) \(89-1\) \(233-1\)
\(=\) \(F_3-1\) \(F_5-1\) \(F_7-1\) \(F_9-1\) \(F_{11}-1\) \(F_{13}-1\)
To see the pattern, notice that each partial sum is one less than a Fibonacci number. Notice that the subscripts are odd. We conjecture
\begin{equation*} \sum_{k=1}^nF_{2k}=F_{2n+1}-1\text{.} \end{equation*}

Subsection 9.3.D Algebra: Factoring Polynomials

In some of our conjectures about sums and products, our answer involved a factored polynomial. We would like to eventually prove these conjectures. For those proofs, it is helpful to be able to factor polynomials. As we saw in Subsection 4.2.B, we use the distributive rule to multiply polynomials. Let’s look at an example using the distributive rule to factor polynomials.

Example 9.3.28 Factoring polynomial.

 
  1. Multiply \((n+s)(n+t)\) and write the answer as a polynomial in the variable \(n\text{.}\)
    Solution.
    Using the distributive property (FOIL) as discussed in Subsection 4.2.B, we get
    \((n+s)(n+t)\) = \(\underbrace{(n)(n)}_{\text{ First } } +\underbrace{(n)(t)}_{\text{ Outside } } + \underbrace{(s)(n)}_{\text{ Inside } } + \underbrace{(s)(t)}_{\text{ Last } }\)
    = \(n^2 + tn + sn + st\)
    = \(n^2+ (s+t)n + st\)
    Observe that the coefficient \(n\) is \(s+t\text{,}\) the sum of \(s\) and \(t\text{,}\) and the constant term is \(st\text{,}\) the product of \(s\) and \(t\text{.}\)
  2. Factor \(n^2 -n-6\text{.}\)
    Solution.
    Based on our observation in (a), we might guess that
    \begin{equation*} n^2 -n-6 = (n+\fillinmath{XXX})(n+\fillinmath{XXX}) \end{equation*}
    where the sum of the integers in the blanks is \(-1\) and the product of the integers in the blanks is \(-6\text{.}\) We guess that the integers are \(-3\) and \(2\) and so
    \begin{equation*} \text{ answer: } n^2 -n-6 = (n-3)(n+2)\text{.} \end{equation*}
    This process of factoring is sometimes nicknamed unfoiling, since we FOIL to check.
    \begin{equation*} \text{ check: } (n-3)(n+2)=n^2 + 2n-3n-6 = n^2-n-6\text{.} \end{equation*}
  3. Multiply \((n+s)(n-s)\) and write the answer as a polynomial in the variable \(n\text{.}\)
    Solution.
    Multiplying we get
    \((n+s)(n-s)\) = \(\underbrace{(n)(n)}_{\text{ First } } +\underbrace{(n)(-s)}_{\text{ Outside } } + \underbrace{(s)(n)}_{\text{ Inside } } + \underbrace{(s)(-s)}_{\text{ Last } }\)
    = \(n^2 - sn + sn - s^2\)
    = \(n^2-s^2\)
  4. Factor \(n^2 -4\text{.}\)
    Solution.
    We can use the result of (c),
    \begin{equation*} n^2-4 = n^2-2^2=(n+2)(n-2) \end{equation*}
Practice using the FOIL method to factor polynomials.

Activity 9.3.D Factoring olynomials.

 
  1. Factor \(m^2+3m+2\text{.}\)
  2. Factor \(m^2-1\text{.}\)
  3. Factor \(2k^2+3k+1\text{.}\) Hint: Try a product of the form \((2k + \underline{})(k+\underline{})\text{.}\)
Let’s look at another example.

Example 9.3.30 Factor quadratic polynomial.

Factor \(3k^2+2k-1\text{.}\)
Solution.
Let’s try a product of the form \((3k + \underline{})(k+\underline{})\text{.}\) To get the constant term \(-1\text{,}\) one of the blanks must be 1 and the other must be \(-1\text{.}\) Since
\begin{equation*} (3k -1)(k+1)=3k^2+3k-k-1=3k^2+2k-1\text{,} \end{equation*}
we can factor \(3k^2+2k-1 = (3k -1)(k+1)\text{.}\)

Subsection 9.3.E Exercises

Subsubsection Exercises for 9.3.A Sum Conjectures

1. Practice.

Calculate the first five partial sums for each sum.
  1. \(\displaystyle \displaystyle \sum_{k=0}^n2 \cdot 3^k\)
  2. \(\displaystyle \displaystyle \sum_{i=1}^{n}3\)
  3. \(\displaystyle \displaystyle \sum_{m=0}^{n}(-1)^m\)
  4. \(\displaystyle \displaystyle \sum_{j=1}^{n}(6j^2)\)
Check.
  1. Hint: \(2, 8, 26, \ldots\)
  2. Hint: \(3, 6, 9, \ldots\)
  3. Hint: \(1, 0, 1, \ldots\)
  4. Hint: \(6, 30, 84, \ldots\)

2. Practice.

 
  1. Use the partial sums from Exercise 1 1.(a) to conjecture an explicit formula for the sum \(\displaystyle \sum_{k=0}^n2 \cdot 3^k\text{.}\)
  2. Use the partial sums from Exercise 1 1.(b) to conjecture an explicit formula for the sum \(\displaystyle \sum_{i=1}^{n}3\text{.}\)
Check.
  1. Hint: The answer involves a power of three, but not exactly \(3^n\)
  2. Hint: The answer is a multiple of 3.

3. Practice.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{k=1}^n(2k-1) \end{equation*}
Check.
Hint: The partial sums are \(1, 4, 9, \ldots\text{.}\)

4. Understand.

 
  1. Use the partial sums from Exercise 1 1.(c) to conjecture an explicit formula for the sum \(\displaystyle \sum_{m=0}^{n}(-1)^m\text{.}\)
  2. Use the partial sums from Exercise 1 1.(d) to conjecture an explicit formula for the sum \(\displaystyle \sum_{j=0}^{n}(6j^2)\text{.}\) Hint: Factor.
Check.
  1. Hint: Use \((-1)^n\) to get a closed formula, noting that \(\frac{1}{2} + \frac{1}{2} =1\) and \(\frac{1}{2} - \frac{1}{2} =0\text{.}\)
  2. Hint: \(6 = (1)(2)(3)\) and \(30 = (2)(3)(5)\text{.}\) The first factor equals \(n\text{,}\) what are the next two factors, in terms of \(n\text{?}\)

5. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \sum_{k=1}^nk(k!)\text{.} \end{equation*}
By the way, the result of this exercise is a key step in proving the uniqueness of the factorial representations introduced in Activity 9.1.B.
Check.
Hint: The partial sums begin \(1, 5, 23, \ldots\) and the formula involves factorials, but not exactly \(n!\text{.}\)

6. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{k=0}^n\frac{1}{2^k} \end{equation*}
Check.
Hint: The partial sums are \(1, \frac{3}{2}, \frac{7}{4}, \ldots\text{.}\) One version of the formula is a fraction in which the numerator involves a power of two (but not exactly \(2^n\)) and the denominator is \(2^n\text{.}\)

7. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{k=0}^n(2k+3) \end{equation*}

8. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum sum \(\displaystyle \sum_{k=1}^n 4 \cdot 5^k\text{.}\)
Check.
Hint: The formula involves powers of five, but not exactly \(5^n\text{.}\)

9. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{k=0}^n(4k+1) \end{equation*}
Check.
Hint: Factor the partial sums.

10. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \sum_{k=1}^n(-1)^kk^2\text{.} \end{equation*}
Check.
Hint: Factor the partial sums.

11. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{k=0}^n(4k-3)\text{.} \end{equation*}
Hint: Factor.

12. Understand.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}\text{.} \end{equation*}
Note that DESMOS adds fractions exactly if you click on the fraction icon.

13. Recap.

Do you know …
  1. How to calculate a partial sum?
  2. Why it is useful to have a recursive definition of a partial sum?
  3. How to use partial sums and pattern recognition to conjecture an explicit formula for a sum?

14. Explore.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum
\begin{equation*} \displaystyle \sum_{m=1}^n6m^3\text{.} \end{equation*}
Hint: Factor.

15. Explore.

Calculate the first few partial sums and use pattern recognition to conjecture an explicit formula for the sum \(\displaystyle \sum_{k=1}^n(-1)^kk\text{.}\) Hint: The formula involves the ceiling.

Subsubsection Exercises for 9.3.B Product Conjectures

16. Practice.

Calculate the first five partial products for each product.
  1. \(\displaystyle \displaystyle \prod_{k=1}^n\frac{1}{k}\)
  2. \(\displaystyle \displaystyle \prod_{i=1}^n3\)
  3. \(\displaystyle \displaystyle \prod_{k=1}^n3k\)
  4. \(\displaystyle \displaystyle \prod_{j=1}^nj^2\)
Check.
  1. Hint: \(\displaystyle 1, \frac{1}{2}, \frac{1}{6}, \ldots\text{.}\)
  2. Hint: \(3, 9, 27, \ldots\text{.}\)
  3. Hint: \(3, 18, 162, \ldots\text{.}\)
  4. Hint: \(1, 4, 36, \ldots\text{.}\)

17. Practice.

 
  1. Use the partial products from Exercise 16 16.(a) to conjecture an explicit formula for the product \(\displaystyle \prod_{k=1}^n\frac{1}{k}\text{.}\)
  2. Use the partial products from Exercise 16 16.(b) to conjecture an explicit formula for the product \(\displaystyle \prod_{i=1}^n3\)
Check.
  1. Hint: The answer involves factorial.
  2. Hint: The answer involves a power of three.

18. Understand.

 
  1. Use the partial products from Exercise 16 16.(c) to conjecture an explicit formula for the product \(\displaystyle \prod_{k=1}^n3k\text{.}\)
  2. Use the partial products from Exercise 16 16.(d) to conjecture an explicit formula for the product \(\displaystyle \prod_{j=1}^nj^2\text{.}\)
Check.
  1. Hint: Keep the partial products factored: \((3 \cdot 1), (3 \cdot 1)(3 \cdot 2),(3 \cdot 1)(3 \cdot 2)(3 \cdot 3)\text{,}\) Then collect the threes.
  2. Hint: Keep the partial products factored: \((1 \cdot 1),(1 \cdot 1)(2 \cdot 2), (1 \cdot 1)(2 \cdot 2)(3 \cdot 3)\text{.}\) The answer involves factorials.

19. Understand.

Calculate the first few partial products and use pattern recognition to conjecture an explicit formula for the product
\begin{equation*} \displaystyle \prod_{j=2}^n \left(1 - \frac{1}{j}\right) \end{equation*}
Check.
Hint: \(1 - \frac{1}{2} = \frac{1}{2}\text{,}\) \(1 - \frac{1}{3} = \frac{2}{3}\text{,}\) \(1 - \frac{1}{4} = \frac{3}{4}\text{,}\) so our product is \(\frac{1}{2}\cdot \frac{2}{3} \cdot \frac{3}{4} \cdots\) and, therefore, the partial product are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots\text{.}\)

20. Understand.

Calculate the first few partial products and use pattern recognition to conjecture an explicit formula for the product \(\displaystyle \prod_{j=1}^n(2j^2)\text{.}\)
Check.
Hint: Keep the partial products factored: \((2\cdot 1 \cdot 1),(2\cdot 1 \cdot 1)(2\cdot 2 \cdot 2),(2\cdot 1 \cdot 1)(2\cdot 2 \cdot 2)(2\cdot 3 \cdot 3) , \ldots\text{.}\) Then collect twos.

21. Understand.

Calculate the first few partial products and use pattern recognition to conjecture an explicit formula for the product \(\displaystyle \prod_{j=1}^n \frac{2}{j}\text{.}\)
Check.
Hint: Keep the numerator and denominator factored and do not reduce the fractions.

22. Recap.

Do you know …
  1. How to calculate a partial product?
  2. Why it is useful to have a recursive definition of a partial product?
  3. How to use partial products and pattern recognition to conjecture an explicit formula for a product?

Subsubsection Exercises for 9.3.C Detour: Fibonacci Sums

23. Practice.

Calculate the first six partial sums and conjecture an explicit form for the sum
\begin{align*} \sum_{k=1}^nF_k \amp = F_1+F_2+F_3+F_4+F_5+ + \ldots +F_n\\ \amp =1 + 1 + 2 + 3 + 5 +\ldots + F_n\text{.} \end{align*}
Check.
Hint: The formula involves Fibonacci numbers. For example, \(12 = 13-1 = F_7-1\text{.}\)

24. Understand.

Calculate the first six partial sums and conjecture an explicit form for the sum
\begin{align*} \sum_{k=1}^n(F_k)^2 \amp = F_1^2+F_2^2+F_3^2+F_4^2+F_5^2+\ldots+F_n^2\\ \amp = 1^2+1^2+2^2+3^2+5^2+\ldots + F_n^2\text{.} \end{align*}
Check.
Hint: The formula involves Fibonacci numbers. Another hint: Factor.

25. Understand.

Calculate the first six partial sums and conjecture an explicit form for the sum
\begin{align*} \sum_{k=1}^{n}F_{2k-1} \amp = F_1+F_3 +F_5 +F_7+F_9+ \ldots + F_{2n-1}\\ \amp = 1+2+5+13+34+\ldots + F_{2n-1}\text{.} \end{align*}
Check.
Hint: the answers involved Fibonacci numbers. For example, \(55 = F_{10}\text{.}\)

26. Recap.

Do you know …
  1. How to calculate a partial sum that involves Fibonacci numbers?
  2. How to use partial sums and pattern recognition to conjecture an explicit formula for a sum involving Fibonacci numbers?

Subsubsection Exercises for 9.3.D Algebra: Factoring Polynomials

28. Understand.

Expand each expression, simplify, then factor the answer as much as possible.
  1. \(\displaystyle (n-1)n + 2n\)
  2. \(\displaystyle n(2n-1)+(4n+1)\)
  3. \(\displaystyle (n-1)n(2n-1) + 6n^2\)

29. Recap.

Do you know …
  1. What is means to factor a polynomial?
  2. How to factor a quadratic polynomial?
  3. How to factor the difference of squares?
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