Free math worksheets

- Apply the recursive update repeatedly to generate the opening terms. - Count how many common-difference steps separate term \(1\) from term \(n\).

a) Starting from \(7\) and adding \(4\) each time gives \(7,11,15,19\). b) From the first term to the \(n\)th term there are \(n-1\) additions of \(4\). Therefore \(a_n=7+4(n-1)=4n+3\).

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) \(a_n=4(-2)^{n-1}\).

- 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 \(n\).

a) The plotted values are \(5,8,11,14,17\), so the first term is \(a_1=5\) and the constant change is \(3\). b) An equivalent recursive definition is \(a_1=5\) and \(a_n=a_{n-1}+3\) for \(n\ge2\). c) There are \(n-1\) increases of \(3\) from the first term to 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) \(a_n=3n+2\).

- 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 \(3,9,81,6561\).

- 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 \(1,3,12,60\).

- 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. - A shifted sequence can turn an affine recurrence into a geometric recurrence. - Check the initial term separately; a recurrence is not complete without its starting value.

a) The first term is \(a_1=6\). Also \(a_n-18=\frac{1}{3}(a_{n-1}-18)\), so \(a_n=\frac{1}{3}a_{n-1}+12\) for \(n\ge2\). b) Let \(b_n=a_n-18\). Then \(b_n=\frac{1}{3}b_{n-1}\) and \(b_1=-12\). Thus \(b_n=-12\left(\frac{1}{3}\right)^{n-1}\), so \(a_n=18-12\left(\frac{1}{3}\right)^{n-1}\).

a) \(a_1=6\), \(a_n=\frac{1}{3}a_{n-1}+12\) for \(n\ge2\). b) Shifting by \(18\) gives the geometric sequence \(b_n=a_n-18=-12\left(\frac{1}{3}\right)^{n-1}\), which recovers the stated 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 \(n\ge2\), \(((n-1)^2+3)+2n-1=n^2+3=a_n\).