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Sequences and Notation: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A centered index

A finite sequence is given by $a_n=|n|+2$ for the integer indices $-2,-1,0,1,2$ . Find $a_0$ .
1. Hint 1
  The subscript is the input to the rule.
2. Hint 2
  The absolute value of zero is zero.
Answer

$a_0=2$ .

Full solution

The index zero is explicitly included, so substitute it into the rule.
$a_0=|0|+2$
Thus $a_0=2$ .
The input is $0$ , while the term it names is $2$ .

Answer

$a_0=2$ .

Key idea

An index identifies a term but need not equal that term's value.
Problem 1: Mark complete
Problem 2 A finite sum

Evaluate $\sum_{k=0}^3(2^k+(-1)^k)$ .
1. Hint 1
  Both limits are included, so there are four summands.
2. Hint 2
  Evaluate the two powers at each of the four integer indices.
Answer

$15$ .

Full solution

The summands at $k=0,1,2,3$ are $2,1,5,7$ , respectively.
$\sum_{k=0}^3(2^k+(-1)^k)=15$
This is four evaluations, including the one at zero.

Answer

$15$ .

Key idea

Sigma notation instructs you to evaluate the entire summand at every included integer index.
Problem 2: Mark complete
Problem 3 Three starting terms

A sequence has $a_1=2$ , $a_2=1$ , and $a_3=4$ . For $n\ge4$ , its rule is $a_n=a_{n-1}+a_{n-3}$ . Find $a_4$ .
1. Hint 1
  Substitute the index four into each subscript of the rule.
2. Hint 2
  The terms required are those at indices $3$ and $1$ .
Answer

$a_4=6$ .

Full solution

At $n=4$ , the recursion requires $a_3$ and $a_1$ , both of which are supplied.
$a_4=a_3+a_1$
Therefore $a_4=4+2=6$ .
The unused seed $a_2$ will be needed when computing $a_5$ .

Answer

$a_4=6$ .

Key idea

A recursion may reach past the immediately preceding term, so its subscripts must be followed carefully.
Problem 3: Mark complete
Problem 4 A plotted list

The graph gives every term of a finite sequence $b_n$ . State its domain, write its five terms in index order, and find $b_2+b_4$ .

The terms of the sequence b.

Text description of this figure

A grid with a horizontal axis labeled n running from -1 to 5 and a vertical axis labeled b running from 0 to 6, on equal unit scales, with a grid line and a label at every integer. Five filled dots stand alone, with no line joining them: above n equals 0 at height 3, above n equals 1 at height 1, above n equals 2 at height 4, above n equals 3 at height 2, and above n equals 4 at height 5. Nothing else is plotted and no dot is labeled.
1. Hint 1
  Read horizontal coordinates as indices and vertical coordinates as term values.
2. Hint 2
  Only the plotted integer inputs belong to this finite sequence.
Answer

Domain $\{0,1,2,3,4\}$ ; terms $3,1,4,2,5$ ; $b_2+b_4=9$ .

Full solution

The plotted horizontal coordinates are $0,1,2,3,4$ , giving the finite domain.
Reading the corresponding heights gives $3,1,4,2,5$ .

The points above indices two and four have values $4$ and $5$ .
$b_2+b_4=9$
The term $b_2$ is the third term in this list because indexing starts at zero.

Answer

Domain $\{0,1,2,3,4\}$ ; terms $3,1,4,2,5$ ; $b_2+b_4=9$ .

Key idea

A sequence graph records separate index-value pairs in the order specified by its domain.
Problem 4: Mark complete
Problem 5 An incomplete rule

A proposed sequence has $a_1=1$ and $a_3=5$ , with $a_n=a_{n-2}+2a_{n-3}$ for $n\ge4$ . Identify the missing seed. Show that choosing it as $1$ or $3$ produces different values of $a_5$ .
1. Hint 1
  A rule beginning at index four does not determine all of its first three terms.
2. Hint 2
  Compute $a_4$ and $a_5$ , tracking which starting terms each one uses.
Answer

The missing seed is $a_2$ ; $a_5=7$ if $a_2=1$ , and $a_5=11$ if $a_2=3$ .

Full solution

The missing initial term is $a_2$ .
The rule gives $a_4=a_2+2a_1$ , which is $a_2+2$ , so even $a_4$ depends on that choice.

At the next step,
$a_5=a_3+2a_2$
which is $5+2a_2$ .
The two proposed seeds give $7$ and $11$ , showing that the incomplete specification does not define one sequence.

Answer

The missing seed is $a_2$ ; $a_5=7$ if $a_2=1$ , and $a_5=11$ if $a_2=3$ .

Key idea

This additive rule starts at index four and reaches back three steps, so all three of its first terms must be given before it defines one sequence.
Problem 5: Mark complete
Problem 6 Two ways to generate

For integers $n\ge0$ , let $c_n=2^n-1$ . Write a recursive rule using only $c_{n-1}$ and constants, including its seed and valid starting index. Find $c_4$ .
1. Hint 1
  An explicit rule and a recursive definition must produce the same term at every allowed index.
2. Hint 2
  Substitute $n-1$ into the explicit formula to express the preceding term.
3. Hint 3
  Use $2^n=2\cdot2^{n-1}$ and compare the constant terms.
Answer

$c_0=0$ ; $c_n=2c_{n-1}+1$ for $n\ge1$ ; $c_4=15$ .

Full solution

The explicit rule gives $c_0=0$ .
Also
$2c_{n-1}+1=2^n-1$
so $c_n=2c_{n-1}+1$ for $n\ge1$ .

The recursion produces $0,1,3,7,15$ through index four.
The explicit check $2^4-1=15$ agrees.

Answer

$c_0=0$ ; $c_n=2c_{n-1}+1$ for $n\ge1$ ; $c_4=15$ .

Key idea

When the next value can be written through the preceding one, as here, an explicit rule becomes a recursion with a seed and a starting index.
Problem 6: Mark complete
Problem 7 A shifted index

Four sensor readings are $r_6=7$ , $r_7=-4$ , $r_8=2$ and $r_9=-8$ . Write their total in sigma notation with index $k$ running from $0$ through $3$ , then evaluate it.
1. Hint 1
  The summation index is a counter and need not equal a reading's subscript.
2. Hint 2
  Choose a subscript expression that gives six at index zero and nine at index three.
Answer

$\sum_{k=0}^3 r_{k+6}=-3$ ; the reversed form $\sum_{k=0}^3 r_{9-k}$ is also valid.

Full solution

The subscript $k+6$ runs through $6,7,8,9$ as $k$ runs from $0$ to $3$ , so the total is
$\sum_{k=0}^3 r_{k+6}.$

Substituting the recorded values gives $7-4+2-8$ , which is $-3$ .
Listing the readings in reverse order, as $\sum_{k=0}^3 r_{9-k}$ , gives the same total.

Answer

$\sum_{k=0}^3 r_{k+6}=-3$ ; the reversed form $\sum_{k=0}^3 r_{9-k}$ is also valid.

Key idea

An index shift lets the summation counter and the recorded subscripts start at different values.
Problem 7: Mark complete
Problem 8 A renaming claim

For a fixed real number $m$ , a student writes $\sum_{k=0}^2(m+k)=\sum_{j=0}^2(m+j)$ . Is that true for every $m$ ? Evaluate both sums in terms of $m$ to justify your answer.
1. Hint 1
  The index changes name but still takes the same three values.
2. Hint 2
  The letter $m$ remains fixed while the temporary index runs through zero, one, and two.
Answer

Yes; both sums equal $3m+3$ .

Full solution

Each sum adds $m$ , $m+1$ , and $m+2$ .
Therefore
$m+(m+1)+(m+2)=3m+3$
The names $k$ and $j$ disappear after evaluation; neither replaces the fixed parameter $m$ .

Answer

Yes; both sums equal $3m+3$ .

Key idea

A fresh summation-index name changes no summands and therefore changes no sum.
Problem 8: Mark complete
Problem 9 A claimed last index

A finite sequence starts at index $4$ and has $7$ terms. A student says its last term is $a_{11}$ . Is that correct? State its last index and explain.
1. Hint 1
  The first index already names one of the terms.
2. Hint 2
  There are six steps between the first and seventh terms.
Answer

No; the last index is $10$ .

Full solution

The seven indices are $4,5,6,7,8,9,10$ .
Equivalently,
$4+(7-1)=10$
Index $11$ would introduce an eighth term, so it is outside the stated finite domain.

Answer

No; the last index is $10$ .

Key idea

A run of k terms spans k minus one steps between its first and last indices.
Problem 9: Mark complete
Problem 10 Repeated values

Let $a_n=n+\frac6n$ for positive integers $n$ . A student claims this cannot be a sequence because $a_2=a_3$ . Is the student correct? Find both terms, state the domain and whether the sequence is finite or infinite, and explain.
1. Hint 1
  A function assigns exactly one output to each allowed input.
2. Hint 2
  Evaluate the two terms, then distinguish different inputs sharing an output from one input receiving different outputs.
Answer

No; $a_2=a_3=5$ . The domain is $\{1,2,3,\ldots\}$ , and the sequence is infinite.

Full solution

At index two, $a_2=2+\frac62$ , which is $5$ , and at index three, $a_3=3+\frac63$ , also $5$ .
Different inputs may share an output; every positive integer still receives exactly one value, so the rule defines a sequence.

The domain contains every positive integer, so the sequence is infinite.

Answer

No; $a_2=a_3=5$ . The domain is $\{1,2,3,\ldots\}$ , and the sequence is infinite.

Key idea

Different indices may name equal terms without violating the definition of a sequence.
Problem 10: Mark complete

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