Question

cc integrated i pdf files should be printed before use. lesson (eng) lección (esp) 8-27. wade and dwayne were working together writing an equation for the sequence 12, 36, 108, 324, ... wade wrote t(n) = 4·3ⁿ and dwayne wrote t(n) = 12·3ⁿ⁻¹. homework help a. make a table for the first four terms of each of their sequences. what do you notice? b. how do you think dwayne explained his method of writing the equation to wade? c. for the sequence 10.3, 11.5, 12.7, ..., wade wrote t(n) = 9.1 + 1.2n while dwayne wrote t(n) = 10.3 + 1.2(n − 1). make a table for the first four terms of each of their sequences. are both forms of the equation correct? d. read the math notes box about standard form of a sequence in this lesson. dwayne’s equations are based on the first term of each sequence, not on the zeroth term. why does dwayne subtract one in both situations? 8.28 on graph paper, draw △abc if a(2, 4), b(0, 5), and c(4, 10) 8-28 hw etool homework help

Explanation:

Part (a)

Step 1: Wade's Formula (\(t(n) = 4 \cdot 3^n\))

For \(n = 1\): \(t(1)=4\cdot3^1 = 12\)
For \(n = 2\): \(t(2)=4\cdot3^2 = 36\)
For \(n = 3\): \(t(3)=4\cdot3^3 = 108\)
For \(n = 4\): \(t(4)=4\cdot3^4 = 324\)

Step 2: Dwayne's Formula (\(t(n) = 12 \cdot 3^{n - 1}\))

For \(n = 1\): \(t(1)=12\cdot3^{0}=12\)
For \(n = 2\): \(t(2)=12\cdot3^{1}=36\)
For \(n = 3\): \(t(3)=12\cdot3^{2}=108\)
For \(n = 4\): \(t(4)=12\cdot3^{3}=324\)

Step 3: Table & Observation

\(n\)	Wade’s \(t(n)\)	Dwayne’s \(t(n)\)
2	36	36
3	108	108
4	324	324

Notice: Both formulas generate the same sequence for \(n = 1,2,3,4\).

Part (b)

Dwayne likely explained that his formula uses the first term (\(a_1 = 12\)) and the common ratio (\(r = 3\)) of the geometric sequence. The standard explicit formula for a geometric sequence is \(t(n)=a_1\cdot r^{n - 1}\), where \(a_1\) is the first term. For the sequence \(12,36,108,324,\dots\), \(a_1 = 12\) and \(r = 3\), so substituting gives \(t(n)=12\cdot3^{n - 1}\). Wade’s formula uses the zeroth term (\(a_0 = 4\), since \(4\cdot3^1 = 12\) for \(n = 1\)), but Dwayne’s formula is structured around the first term (\(n = 1\) gives the first term directly, as \(3^{0}=1\)).

Part (c)

Step 1: Wade's Formula (\(t(n)=9.1 + 1.2n\))

For \(n = 1\): \(t(1)=9.1+1.2(1)=10.3\)
For \(n = 2\): \(t(2)=9.1+1.2(2)=11.5\)
For \(n = 3\): \(t(3)=9.1+1.2(3)=12.7\)
For \(n = 4\): \(t(4)=9.1+1.2(4)=13.9\)

Step 2: Dwayne's Formula (\(t(n)=10.3 + 1.2(n - 1)\))

For \(n = 1\): \(t(1)=10.3+1.2(0)=10.3\)
For \(n = 2\): \(t(2)=10.3+1.2(1)=11.5\)
For \(n = 3\): \(t(3)=10.3+1.2(2)=12.7\)
For \(n = 4\): \(t(4)=10.3+1.2(3)=13.9\)

Step 3: Table & Check

\(n\)	Wade’s \(t(n)\)	Dwayne’s \(t(n)\)
2	11.5	11.5
3	12.7	12.7
4	13.9	13.9

Both equations are correct because they generate the same arithmetic sequence (\(10.3,11.5,12.7,13.9,\dots\)) for \(n = 1,2,3,4\). Wade’s formula uses a “zeroth term” (\(9.1\) when \(n = 0\): \(9.1+1.2(0)=9.1\)), while Dwayne’s uses the first term (\(10.3\) when \(n = 1\)) with the standard arithmetic sequence formula \(t(n)=a_1 + d(n - 1)\) (where \(a_1 = 10.3\), \(d = 1.2\)).

Part (d)

Answer:

Dwayne subtracts 1 (\(n - 1\)) to align the formula with the first term of the sequence (\(n = 1\)). For a geometric sequence, the standard explicit formula is \(t(n)=a_1\cdot r^{n - 1}\) (so \(n = 1\) gives \(a_1\), since \(r^{0}=1\)). For an arithmetic sequence, the standard explicit formula is \(t(n)=a_1 + d(n - 1)\) (so \(n = 1\) gives \(a_1\), since \(d(0)=0\)). By subtracting 1, Dwayne ensures \(n = 1\) directly produces the first term of the sequence, rather than a “zeroth term” (used in Wade’s formulas, where \(n = 0\) gives the initial value before the first term). This follows the standard form for sequence equations, where the index \(n\) starts at 1 for the first term.