- Check the increase from the first term to each later term. How does it change as the term number increases? - How many additions separate the first term from the \(n\)th term? - Test your explicit formula at the first and fourth terms.

1. Starting at \(7\), add \(4\) each time to obtain \(7,11,15,19\). 2. There are \(n-1\) additions from the first term to the \(n\)th term, so \(a_n=7+4(n-1)=4n+3\).

a) \(7,11,15,19\). b) \(a_n=4n+3\).

- Evaluate the explicit rule at its starting index first. - Compare the explicit values at consecutive indices to identify the constant update. - A recursive definition needs both an initial value and a rule for later terms.

1. The first term is \(a_1=20\). 2. The explicit formula decreases by \(3\) whenever \(n\) increases by \(1\), so \(a_n=a_{n-1}-3\) for \(n\ge2\). 3. The first three terms are \(20,17,14\), and both definitions produce them.

\(a_1=20\), \(a_n=a_{n-1}-3\) for \(n\ge2\). The first three terms are \(20,17,14\).

- Apply the recursive multiplier carefully, including its sign, to generate a few terms. - Look for the repeated multiplicative factor in the generated terms. - In an explicit geometric form, the exponent should be \(0\) at the starting index.

1. Repeatedly applying the recurrence gives \(4,-8,16,-32\). 2. Each term is obtained by multiplying the preceding term by \(-2\), so the sequence is geometric with first term \(4\) and ratio \(-2\). 3. Therefore the explicit form is \(a_n=4(-2)^{n-1}\).

a) \(4,-8,16,-32\). b) \(a_n=4(-2)^{n-1}\).

- Apply the same addition rule separately to each starting value. - If the starting values were hidden, could the update rule tell you which list to produce?

1. Starting at \(2\) gives \(2,5,8\). 2. Starting at \(5\) gives \(5,8,11\). 3. The same update rule produces different sequences when the starting value changes, so an initial condition is needed.

A: \(2,5,8\). B: \(5,8,11\). The update rule alone does not specify the initial value.

- Substitute the starting index into the explicit formula. - Compare what happens to the value when the exponent increases by one.

1. At \(n=0\), \(a_0=16\). Each term is half the previous term, so \(a_{n+1}=\frac{1}{2}a_n\) for \(n\ge0\). 2. The first terms are \(16,8,4,2\), so \(a_3=2\).

a) \(a_0=16\), \(a_{n+1}=\frac{1}{2}a_n\) for \(n\ge0\). b) \(a_3=2\).

- Evaluate the explicit formula at its first allowed index. - Compare the powers in two consecutive explicit terms to find the multiplicative update. - Include the starting value because the recursive rule alone does not determine a unique sequence.

1. The initial term is \(a_1=6\left(-\frac{1}{2}\right)^0=6\). 2. Each time \(n\) increases by \(1\), the explicit expression gains one factor of \(-\frac{1}{2}\). 3. Therefore, \(a_n=-\frac{1}{2}a_{n-1}\) for \(n\ge2\).

\(a_1=6\) and \(a_n=-\frac{1}{2}a_{n-1}\) for \(n\ge2\).

- Generate a few terms while keeping their indices visible. - Count the number of recursive multiplications from index \(3\) to a general index \(n\). - Check that the exponent in the explicit form becomes \(0\) at the starting index.

1. Repeatedly doubling gives \(a_3=5\), \(a_4=10\), \(a_5=20\), and \(a_6=40\). 2. The term \(a_n\) is \(n-3\) doubling steps after \(a_3\). 3. Therefore, \(a_n=5\cdot2^{n-3}\) for \(n\ge3\).

a) \(5,10,20,40\). b) \(a_n=5\cdot2^{n-3}\) for \(n\ge3\).

- Read the plotted points from left to right and compare consecutive vertical values. - A recursive form needs the starting value shown by the first plotted point and a rule for moving to the next point. - For the explicit form, count how many equal changes occur between index \(1\) and index \(n\).

1. The plotted terms are \(5,8,11,14,17\). The first is \(a_1=5\), and the constant change is \(3\). 2. A recursive definition is \(a_1=5\) and \(a_n=a_{n-1}+3\) for \(n\ge2\). 3. There are \(n-1\) increases of \(3\) by term \(n\), so \(a_n=5+3(n-1)=3n+2\).

a) \(a_1=5\) and the constant change is \(3\). b) \(a_1=5\), \(a_n=a_{n-1}+3\) for \(n\ge2\). c) \(a_n=3n+2\).

- Start by evaluating the explicit formula at the initial index. - Compare \(a_n\) with \(a_{n-1}\) by taking their ratio. - Use the factorial cancellation to identify the factor that changes with \(n\).

1. The initial term is \(a_0=\frac{2!}{2}=1\). 2. For \(n\ge1\), \(\frac{a_n}{a_{n-1}}=\frac{(n+2)!/2}{(n+1)!/2}=n+2\). 3. Therefore \(a_n=(n+2)a_{n-1}\) for \(n\ge1\). 4. The first four terms are \(1,3,12,60\), and both definitions produce those values.

\(a_0=1\), \(a_n=(n+2)a_{n-1}\) for \(n\ge1\). The terms for \(n=0,1,2,3\) are \(1,3,12,60\).

- Evaluate the explicit formula at the first index it allows. - Keep the constant change while checking whether the initial term agrees. - Verify more than one term after making the correction.

1. The explicit formula gives \(a_1=2(1)+1=3\), not \(1\). 2. The common difference is \(2\), so the correct recurrence is \(a_1=3\), \(a_n=a_{n-1}+2\) for \(n\ge2\). 3. Both formulas give \(a_1=3\), \(a_2=5\), and \(a_3=7\).

The starting value is wrong. Use \(a_1=3\), \(a_n=a_{n-1}+2\) for \(n\ge2\). Both forms give \(3,5,7\) for the first three terms.

- Compare the formulas at indices \(n\) and \(n+1\). - Remember that the added amount need not be constant. - Check the first index separately before using the update rule.

1. Subtract successive formulas: \(a_{n+1}-a_n=((n+1)^2+2)-(n^2+2)=2n+1\). 2. At \(n=0\), \(a_0=2\), so \(a_{n+1}=a_n+2n+1\) for \(n\ge0\). 3. Both forms give \(a_0=2\), \(a_1=3\), \(a_2=6\), and \(a_3=11\).

a) \(a_{n+1}-a_n=2n+1\). b) \(a_0=2\), \(a_{n+1}=a_n+2n+1\); the first four terms are \(2,3,6,11\).

- Decide how many doublings separate the first bed from bed \(n\). - A recursive formula needs a starting value and a next-term rule. - Check one later bed by updating repeatedly and by direct substitution.

1. The starting value is \(a_1=4\). Doubling gives \(a_n=2a_{n-1}\) for \(n\ge2\). 2. Bed \(n\) is obtained after \(n-1\) doublings, so \(a_n=4\cdot2^{n-1}\). 3. Recursively the first four values are \(4,8,16,32\). Explicitly, \(a_4=4\cdot2^3=32\).

a) Recursive: \(a_1=4\), \(a_n=2a_{n-1}\) for \(n\ge2\). Explicit: \(a_n=4\cdot2^{n-1}\). b) Both give \(a_4=32\) seedlings.

- Recover an individual term from two consecutive partial sums. - After finding the explicit form, compare the expressions for consecutive terms. - A recursive definition needs both a starting value and a rule that produces each later term from the preceding one.

1. Since \(a_1=S_1=5\), and for \(n\ge2\), \(a_n=S_n-S_{n-1}\) \(=(2n^2+3n)-[2(n-1)^2+3(n-1)]=4n+1\). Thus the explicit form is \(a_n=4n+1\) for \(n\ge1\). 2. The first term is \(a_1=5\). Consecutive explicit terms differ by \(a_n-a_{n-1}=(4n+1)-[4(n-1)+1]=4\). Therefore an equivalent recursive definition is \(a_1=5\) and \(a_n=a_{n-1}+4\) for \(n\ge2\). 3. The explicit form increases by \(4\) whenever \(n\) increases by \(1\), exactly matching the recursive update.

1) \(a_n=4n+1\). 2) \(a_1=5\) and \(a_n=a_{n-1}+4\) for \(n\ge2\). 3) The difference between consecutive explicit terms is \(4\), so the two forms describe the same sequence.

- Use the known indexed term in the explicit form before working with the recursive parameters. - The recursive starting value must match the explicit formula at index \(2\). - Compare two consecutive explicit terms to determine the recursive change. - Verify equivalence symbolically rather than checking only a few numerical terms.

1. Use \(a_7=32\) in the explicit form: \(32=5\cdot7+c\), so \(c=-3\). 2. Then \(a_2=5\cdot2-3=7\), so \(p=7\). 3. Consecutive explicit terms differ by \([5n-3]-[5(n-1)-3]=5\), so \(d=5\). 4. With these values, the recursive form is \(a_2=7\), \(a_n=a_{n-1}+5\). If \(a_{n-1}=5(n-1)-3\), then \(a_n=5(n-1)-3+5=5n-3\), which is exactly the explicit form. Thus the two descriptions agree for every \(n\ge3\).

\(p=7\), \(d=5\), and \(c=-3\). The recursion \(a_n=a_{n-1}+5\) advances the explicit expression \(5(n-1)-3\) to \(5n-3\).

- Compare how the difference between \(a_n\) and the constant part of the explicit formula changes from one index to the next. - After choosing a value to subtract, test the first three shifted terms. What should their consecutive ratios show? - Check the initial term separately; a recurrence is not complete without its starting value.

1. The first term is \(a_1=6\). Since \(a_n-18=\frac{1}{3}(a_{n-1}-18)\), we get \(a_n=\frac{1}{3}a_{n-1}+12\) for \(n\ge2\). 2. Let \(b_n=a_n-18\). Then \(b_n=\frac{1}{3}b_{n-1}\) and \(b_1=-12\), so \(b_n=-12\left(\frac{1}{3}\right)^{n-1}\). 3. Adding \(18\) gives \(a_n=18-12\left(\frac{1}{3}\right)^{n-1}\), as required.

a) \(a_1=6\), \(a_n=\frac{1}{3}a_{n-1}+12\) for \(n\ge2\). b) Shifting by \(18\) gives \(b_n=a_n-18=-12\left(\frac{1}{3}\right)^{n-1}\), recovering the explicit form.

- Describe the \(n\)th term as the initial value plus every increment added from index \(2\) through index \(n\). - Simplify the accumulated odd-number sum before writing the explicit rule. - A general verification must check the starting value and show that one recursive update reproduces the explicit expression for an arbitrary \(n\).

1. Accumulating the changes from index \(2\) through index \(n\) gives \(a_n=4+\sum_{k=2}^{n}(2k-1)\). 2. Since \(\sum_{k=1}^{n}(2k-1)=n^2\), the sum from \(k=2\) to \(n\) is \(n^2-1\). 3. Therefore \(a_n=n^2+3\). 4. The initial value checks because \(1^2+3=4\). 5. For \(n\ge2\), substituting the explicit formula for the previous term gives \(a_{n-1}+2n-1=((n-1)^2+3)+2n-1=n^2+3=a_n\). Thus the explicit formula satisfies the recursion for every allowed \(n\).

\(a_n=n^2+3\). It has \(a_1=4\), and for every \(n\ge2\), \(((n-1)^2+3)+2n-1=n^2+3=a_n\).

- Compare the exponent \(2^n\) with the exponent at the preceding index. - Think about what operation on an entire power doubles its exponent. - Include the value at \(n=0\) as the starting term of the recursion.

1. The initial term is \(a_0=3\). 2. Since \(2^n=2\cdot2^{n-1}\), \(a_n=3^{2\cdot2^{n-1}}=\left(3^{2^{n-1}}\right)^2\). 3. Therefore \(a_n=(a_{n-1})^2\) for \(n\ge1\). 4. Both forms give \(3,9,81,6561\) for \(n=0,1,2,3\).

\(a_0=3\), \(a_n=(a_{n-1})^2\) for \(n\ge1\). The first four terms are \(3,9,81,6561\).