2. Representing Patterns (A)
  3. Lesson 8: Equivalent Expressions for the General Term

Answers and Solutions


    1. Using a table, we can test each expression by substituting some values of \(n\).
      \(n\) \(5n + 3\) \(7n + 2 - 2n\)
      \(1\) \(8\) \(7\)
      \(2\) \(13\) \(12\)
      Since the values of the expressions are not equal for \(n=1\) and \(n=2\), we can conclude that \(5n + 3\) is not equivalent to \(7n + 2 - 2n\).
    2. Using a table, we can test each expression by substituting some values of \(n\).
      \(n\) \(9n-4\) \(5n\)
      \(1\) \(5\) \(5\)
      \(2\) \(14\) \(10\)
      Since the values of the expressions are not equal for \(n=2\), we can conclude that \(9n-4\) is not equivalent to \(5n\).
    3. Using a table, we can test each expression by substituting some values of \(n\).
      \(n\) \(4(n+6)\) \(4n + 24\)
      \(1\) \(28\) \(28\)
      \(2\) \(32\) \(32\)
      Since the values of the expressions are equal for \(n=1\) and \(n=2\), we should try to show that \(4(n+6)\) and \(4n + 24\) are equivalent.
      Recall that multiplication distributes over addition.  For example: \[\begin{align*} 4 \times (1 + 6) &= 4 \times 1 + 4 \times 6 = 4 \times 1 + 24\\ 4 \times (2 + 6) & = 4 \times 2 + 4 \times 6 = 4 \times 2 + 24\\ 4 \times (3 + 6) & = 4 \times 3 + 4 \times 6 = 4 \times 3 + 24 \end{align*}\]

      We can use this distribution property to explain that  \(4\times (n + 6) = 4 \times n + 4\times 6 = 4 n + 24\) no matter what value of \(n\) we choose.  Therefore, the two expressions are equivalent.

  2. Notice in Image 1, there are \(2\) vertical columns of red squares joined by \(1\) blue square at the base. The number of red squares is represented by the expression \(2(n+1)\) because there are \(2\) groups with \(n+1\) squares in each.  If we add the number of red squares and the number of blue squares, we get the expression \(2(n+1) +1\)which describes the total number of squares in each term of the image sequence.

    In Image 2, there are \(2\) vertical columns of red squares and \(3\) blue squares forming the base. The number of red squares is represented by the expression \(2n\) because there are \(2\) groups with \(n\) squares in each. The expression \(2n+3\) describes the total number of squares in each term of the image sequence.
    1. Total Number of Toothpicks
      The sequence \(5,~9,~13, \ldots\) represents the total number of toothpicks in each term of the image sequence. The general term of this sequence is \(4n + 1\).

      Break It Down
      Term 1 contains 5 toothpicks arranged to form a house shape, where the square base is grouped and the triangular peaks are another group.Term 2 contains 9 toothpicks arranged to form two house shapes with an adjoining side, where the square bases is grouped and the triangular peaks are another group.Term 3 contains 13 toothpicks arranged to form three house shapes with two adjoining sides, where the square bases is grouped and the triangular peaks are another group.
      The sequence \(\class{hl1}{3,~5,~7, \ldots}\) represents the number of toothpicks that make up the square bases. The general term of this sequence is \(\class{hl1}{2n+1}\).
      The sequence \(\class{hl2}{2,~4,~6, \ldots}\) represents the number of toothpicks that compose the triangular peaks. The general term of this sequence is \(\class{hl2}{2n}\).
      Therefore, the general term representing all the toothpicks in each term of the image sequence is \(\class{hl1}{2n+1}+\class{hl2}{2n}\).

      Add On To Term 1
      Notice that the first image uses \(5\) toothpicks and then exactly \(4\) toothpicks are added each time to obtain every term after the first. Therefore, the sequence \(5,~5+4,~5+4+4, \ldots\) represents the total number of toothpicks in each term of the image sequence. The general term of this sequence is \(5+4(n-1)\).
    2. Total Number of Toothpicks
      The sequence \(6,~11,~16, \ldots\) represents the total number of toothpicks in each term of the image sequence. The general term of this sequence is \(5n + 1\).

      Break It Down
      Term 1 contains 6 toothpicks to form a hexagon, where the vertical toothpicks are grouped and the slanted are another group.Term 3 contains 16 toothpicks arranged to form three hexagons with two adjoining sides, where the vertical toothpicks are grouped and the slanted are another group.
      The sequence \(\class{hl1}{2,~3,~4, \ldots}\) represents the number of vertical toothpicks. The general term of this sequence is \(\class{hl1}{n+1}\).
      The sequence \(\class{hl2}{4,~8,~12, \ldots}\) represents the number of slanted toothpicks. The general term of this sequence is \(\class{hl2}{4n}\).
      Therefore, the general term representing all the toothpicks in each term of the image sequence is \(\class{hl1}{n+1}+\class{hl2}{4n}\).

      Add On To Term 1
      Notice that the first image uses \(6\) toothpicks and then exactly \(5\) toothpicks are added each time to obtain every term after the first. The sequence \(6,~6+5,~6+5+5, \ldots\) represents the total number of toothpicks in each term of the image sequence. The general term of this sequence is \(6+5(n-1)\).
    3. Total Number of Toothpicks
      The sequence \(6,~9,~12, \ldots\) represents the total number of toothpicks in each term of the image sequence. The general term of this sequence is \(3n + 3\).

      Break It Down 
      Term 1 contains 4 toothpicks that form a square, and 2 additional toothpicks to form a triangle on the right adjoining side. The square base is grouped and the triangular peaks are another group.Term 2 contains 7 toothpicks that form two adjoining squares, and 2 additional toothpicks to form a triangle on the right adjoining side. The square base is grouped and the triangular peaks are another group.Term 3 contains 10 toothpicks that form three adjoining squares, and 2 additional toothpicks to form a triangle on the right adjoining side. The square base is grouped and the triangular peaks are another group.
      The sequence \(\class{hl1}{4,~7,~10, \ldots}\) represents the number of toothpicks that make up the squares. The general term of this sequence is \(\class{hl1}{3n+1}\).
      The sequence \(\class{hl2}{2,~2,~2, \ldots}\) represents the number of toothpicks that compose the triangular peaks. The general term of this sequence is \(\class{hl2}{2}\).
      Therefore, the general term representing all the toothpicks in each term of the image sequence is \(\class{hl1}{3n+1}+\class{hl2}{2}\).  (This we should recognize as the same as the expression above.)

      Add On To Term 1
      Notice that the first image uses \(6\) toothpicks and then exactly \(3\) toothpicks are added each time to obtain every term after the first. The sequence \(6,~6+3,~6+3+3, \ldots\) represents the total number of toothpicks in each term of the image sequence. The general term of this sequence is \(6+3(n-1)\).
  4. Answers may vary.
    The following image represents the expression \((2n+6)+n\):
    Term 1 contains 3 by 3 squares with the middle square dotted and the rest shaded. Term 2 contains 4 by 3 squares with the 2 middle squares dotted and the rest shaded. Term 3 contains 5 by 3 squares, with the 3 middle squares dotted and the rest shaded.
    The following image represents the expression \(2(n+2) + (n+2)\):
    Term 1 contains 3 by 3 squares with the top row dotted and the rest shaded. Term 2 contains 4 by 3 squares with the top row dotted and the rest shaded. Term 3 contains 5 by 3 squares, with the top row dotted and the rest shaded.
    1. To determine whether Lesia and Devan have the same sequence, we can use a table to find some terms in each sequence.
      Term Number (\(n\)) Lesia's Sequence Devan's Sequence
      \(1\) \(7\) \(6\)
      \(2\) \(12\) \(11\)
      \(3\) \(17\) \(16\)
      \(4\) \(22\) \(21\)
      We can see from the table that the terms in the two sequences are not the same and therefore Lesia and Devan must be describing different sequences.
    2. When we look at the table, we notice that Lesia and Devan's sequences are similar in that consecutive terms in each sequence all differ by \(5\). The result is that each term in Devan's sequence is \(1\) less than the corresponding term in Lesia's sequence.  Can you see this relationship by comparing the two general terms?  We could also represent Devan's sequence using the general term \(5n+1\) which would always produce a number one less than the general term \(5n+2\).
    1. If \(n\) represents the term number of the value you are trying to find, then \(n-1\) represents the one less than the term number. To find the \(n^{th}\) term in this sequence, subtract the square of \(n-1\) from the square of \(n\).
      To calculate the values for the first eight terms in this sequence, we can use a table to organize our work.
      Term Number Term in Sequence Term Value
      \(1\) \(1^2-0^2\) \(1\)
      \(2\) \(2^2-1^2\) \(3\)
      \(3\) \(3^2-2^2\) \(5\)
      \(4\) \(4^2-3^2\) \(7\)
      \(5\) \(5^2-4^2\) \(9\)
      \(6\) \(6^2-5^2\) \(11\)
      \(7\) \(7^2-6^2\) \(13\)
      \(8\) \(8^2-7^2\) \(15\)
    2. Using the last two columns, we can view the same sequence in two different ways, which will help us to come up with two expressions to represent term \(n\) in the given sequence. In the second column, we can compare the bases of both powers to the term number. When we do, we determine that the expression\[n^2-(n-1)^2\] represents term \(n\) of the sequence. In the third column, we notice that the difference between consecutive terms is \(2\), and using our previous strategies we can determine that\[2n-1\] also represents term \(n\) of the sequence. 
    1. First we recall that the letter \(n\) represents an unknown term number. One way that we can try to show that these two expressions represent the same sequence of numbers is to model the expressions using a diagram.
      Let ★ (a star) represent \(n\) and let ⬥ (a diamond) represent \(1\). To represent the expression \(7n-3n+1+2\), we must:
      Step 1: Draw \(7\) stars to represent \(7n\).
      ★ ★ ★ ★ ★ ★ ★
      Step 2: Remove (or cross out) \(3\) stars to represent the subtraction of \(3n\).
      ★ ★ ★ ★ ★ ★ ★
      Step 3: Add \(1\) diamond to represent \(+1\).
      ★ ★ ★ ★ ⬥
      Step 4: Add \(2\) diamonds to represent \(+2\).
      ★ ★ ★ ★ ⬥ ⬥ ⬥ 
      What do we have left after all of these steps? 
      We have \(4\) stars which represents \(4n\) and we have \(3\) diamonds which represents \(3\). Therefore, not only does our diagram represent the expression \(7n-3n+1+2\), it also represents the expression \(4n+3\). Since these expressions are equivalent, we know that \(4n+3\) must also represent term \(n\) of this sequence.
    2. Answers may vary. We could use the same thinking as above with diagrams, or we can work directly with the variables and the numbers. In the expression \(3n-5\), we notice that there are two different parts to this expression, the \(3n\) and the \(-5\). It makes sense that we can manipulate each part separately to produce new expressions.
      For example, \(-5=-2-3\) and so the expression \(3n-2-3\) is equivalent to the expression \(3n-5\).
      Similarly, \(3n = n+2n\) and so the expression \(n+2n-5\) is equivalent to the expression \(3n-5\).
      Once you get the hang of changing the look of each term, you can come up with many different expressions that are equivalent to \(3n-5\). The following are a few more examples of some equivalent expressions:\[n+2 + 2n - 7 \qquad 5n-2n -9 + 4 \qquad -4n+7n +10-15\] What is the most interesting expression that you can find?
    1. To verify that these two expressions give the same value for \(n=1,2,3,4,5\), we can use a table.
      \(n\) \((n+1)^2-n^2\) \(2n+1\)
      \(1\) \( \begin{align*} (n+1)^2-n^2 &= (1+1)^2 - 1^2 \\ &= 2^2 - 1^2 \\ &= 3 \end{align*}\) \(\begin{align*} 2n+1 &= 2(1) +1 \\ &= 3 \end{align*}\)
      \(2\) \( \begin{align*} (n+1)^2-n^2 &= (2+1)^2 - 2^2 \\ &= 3^2 - 2^2 \\ &= 5 \end{align*}\) \(\begin{align*} 2n+1 &= 2(2) +1 \\ &= 5 \end{align*}\)
      \(3\) \( \begin{align*} (n+1)^2-n^2 &= (3+1)^2 - 3^2 \\ &= 4^2 - 3^2 \\ &= 7 \end{align*}\) \(\begin{align*} 2n+1 &= 2(3) +1 \\ &= 7 \end{align*}\)
      \(4\) \( \begin{align*} (n+1)^2-n^2 &= (4+1)^2 - 4^2 \\ &= 5^2 - 4^2 \\ &= 9 \end{align*}\) \(\begin{align*} 2n+1 &= 2(4) +1 \\ &= 9 \end{align*}\)
      \(5\) \( \begin{align*} (n+1)^2-n^2 &= (5+1)^2 - 5^2 \\ &= 6^2 - 5^2 \\ &= 11 \end{align*}\) \(\begin{align*} 2n+1 &= 2(5) +1 \\ &= 11 \end{align*}\)
      Comparing the values in each row, we have verified that these two expressions give the same value for \(n=1,~2,~3,~4,~5\).
    2. Since the expressions are equivalent for \(n=1,~2,~3,~4,~5\), it makes sense to check if these two expressions might be equivalent.  
      These expressions are very similar to the expressions that you worked with in an earlier problem in this problem set. Also, you may have already seen an exercise in this course that uses diagrams to explore the positive difference between consecutive square numbers. To form a solution, we consider the following coloured diagrams of larger squares formed by arranging unit squares.

      Observe the following: Inside the square with side length \(2\), there is a square of side length \(1\), inside the square of side length \(3\) there is a square of side length \(2\), and so on.  
      Let's break down the diagrams even further as follows:

      Looking at the \(5 \times 5\) square, we see that the difference in area between a \(5\times 5\) square and a \(4 \times 4\) square is exactly the area of \(2 \times 4 + 1 = 9\) unit squares.  More generally, the difference between the area of an \((n+1)\times (n+1)\) square and the area of an \(n \times n\) square is exactly the area of \(2\times n + 1\)unit squares.  Since both of the expressions \[(n+1)^{2} - n^{2}\text{ and } 2n +1\]

      represent the same area, they must be equivalent!