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.

The first term is \(u_1=20\). The explicit formula decreases by \(3\) whenever \(n\) increases by \(1\), so \(u_n=u_{n-1}-3\) for \(n\ge2\). The first three terms are \(20,17,14\), and both definitions produce them.

\(u_1=20\), \(u_n=u_{n-1}-3\) for \(n\ge2\). The first three terms are \(20,17,14\).

- 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.

The initial term is \(b_1=6\left(-\frac12\right)^0=6\). Each time \(n\) increases by \(1\), the explicit expression gains one factor of \(-\frac12\). Therefore \(b_n=-\frac12b_{n-1}\) for \(n\ge2\).

\(b_1=6\) and \(b_n=-\frac12b_{n-1}\) 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.

a) Repeatedly doubling gives \(c_3=5\), \(c_4=10\), \(c_5=20\), and \(c_6=40\). b) The term \(c_n\) is \(n-3\) doubling steps after \(c_3\). Therefore \(c_n=5\cdot2^{n-3}\) for \(n\ge3\).

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

- Express \(a_n\) as the starting value plus all the successive changes from index \(2\) through index \(n\). - The added amount is a linear expression in the index, so its cumulative sum will not remain linear. - Verify an explicit rule by checking that it matches both the starting term and several recursive updates.

1. Accumulating the changes 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 recursion gives \(4,7,12,19\) for \(n=1,2,3,4\), and the explicit formula gives \(1^2+3=4\), \(2^2+3=7\), \(3^2+3=12\), and \(4^2+3=19\).

\(a_n=n^2+3\).