Explicit & Recursive Formulas - Algebra 3

An arithmetic sequence is defined by $a_n=2n+1$ for $n \ge 1$ . Which recursive formula using $a_{n+1}$ also defines the sequence?
Answer: $a_1=3, a_{n+1}=a_n+2$ for $n \ge 1$ Whenever you see an explicit formula for an arithmetic sequence, your task is to extract the starting value and the common difference. The explicit form $a_n = a_1 + (n-1)d$ has the same structure as $a_n = 2n+1$ . Rewriting $2n+1$ as $3 + (n-1)2$ shows that $a_1=3$ and the common difference is $d=2$ . So the recursive definition must start at 3 and add 2 each time: $a_1=3,\ a_{n+1}=a_n+2$ . The choice starting with $a_1=1$ misidentifies the first term—plugging $n=1$ into $2n+1$ gives 3, not 1. The choice with $a_{n+1}=a_n-2$ starts correctly at 3 but subtracts 2, producing $3,1,-1,\dots$ , a different arithmetic sequence. The choice with $a_{n+1}=2a_n$ also starts at 3 but multiplies each term by 2, giving $3,6,12,\dots$ , which is geometric, not the linear pattern $3,5,7,\dots$ we need. A useful strategy: compute the first two terms from the explicit formula, then check which recursive rule generates those same values from the same starting point. That quickly eliminates all traps and confirms the rule.
A sequence satisfies $a_{n+1}=a_n-5$ for $n \ge 1$ , and $a_3=12$ . Which explicit formula gives $a_n$ for $n \ge 1$ ?
Answer: $a_n=27-5n$ Whenever you see a sequence like $a_{n+1}=a_n-5$ , recognize that this is an arithmetic sequence with common difference $d=-5$ . Your job is to shift from the recursive form to an explicit formula using any known term. Since $a_3=12$ , you can anchor the formula at $n=3$ : $a_n=a_3+(n-3)(-5)=12-5n+15=27-5n.$ So the correct explicit formula is $a_n=27-5n$ . Quick check: at $n=3$ , it gives $27-15=12$ . The choice $17-5n$ comes from treating $a_3=12$ as if it were $a_1$ : that would mean $12-5(n-1)=17-5n$ , but it makes $a_3=2$ , not $12$ . The choice $12-5n$ simply attaches $-5n$ to the given term, ignoring that $n$ must count steps from $n=3$ ; it gives $-3$ when $n=3$ . The choice $22-5n$ finds $a_1=22$ correctly but then forgets the $(n-1)$ adjustment: the correct form is $22-5(n-1)=27-5n$ , not $22-5n$ . Your strategy: anchor the explicit rule at the known term, using $a_n=a_k+(n-k)d$ . This prevents both off-by-one mistakes and the trap of plugging the given term into the wrong position.
A geometric sequence has $a_4=24$ and $a_7=192$ . Which explicit formula gives $a_n$ for $n \ge 1$ ?
Answer: $a_n=3(2)^{n-1}$ This question tests whether you can find the common ratio and translate it into the explicit form of a geometric sequence: $a_n=a_1 r^{n-1}$ . Since the terms are four apart? Actually $a_7$ is three steps after $a_4$ , so $a_7=a_4 r^3$ . Substitute: $192=24r^3$ , so $r^3=8$ and $r=2$ . Then use $a_4=24$ : $24=a_1(2)^3=8a_1$ , so $a_1=3$ . Therefore, $a_n=3(2)^{n-1}$ . Check: at $n=4$ , $3(2)^3=24$ ; at $n=7$ , $3(2)^6=192$ . The choice $3(2)^n$ and the choice $6(2)^{n-1}$ are actually the same sequence; both are twice the correct formula, giving $a_4=48$ instead of 24. They trap students who write $a_1 r^n$ instead of $a_1 r^{n-1}$ . The choice $24(2)^{n-3}$ treats 24 as if it were $a_1$ , but 24 is $a_4$ ; at $n=4$ it produces 48, not 24. If you anchor at $a_4$ , the correct form would be $24(2)^{n-4}$ , which simplifies to $3(2)^{n-1}$ . Study tip: always plug the given term numbers into any formula you choose. If both known values match, the formula is correct. And remember: for a sequence starting at $n=1$ , the exponent is $n-1$ , not $n$ .
The sequence is defined by $a_1 = 8$ and $a_n = a_{n-1} - \frac{1}{2}$ for $n \ge 2$ . Which explicit formula represents the same sequence?
Answer: $a_n=8-\frac{1}{2}(n-1)$ Whenever you see a recursive definition like $a_n=a_{n-1}-\frac12$ , recognize it as an arithmetic sequence: each term is made by adding the same common difference, here $d=-\frac12$ . The explicit formula for an arithmetic sequence is $a_n=a_1+(n-1)d$ , so substituting $a_1=8$ and $d=-\frac12$ gives $a_n=8-\frac12(n-1).$ That formula correctly produces $a_1=8$ , then $a_2=7.5$ , matching the recursion. The choice $8-\frac12 n$ is a classic off-by-one error: it uses $nd$ instead of $(n-1)d$ , so at $n=1$ it gives $7.5$ , not $8$ . The choice $8+\frac12(n-1)$ uses the wrong sign for the common difference; it describes a sequence that increases by $0.5$ , the opposite direction of the given rule. The choice $\frac{15}{2}-\frac12 n$ has the correct slope but the wrong starting point: at $n=1$ it gives $7$ , not $8$ , so it represents a different arithmetic sequence entirely. A fast way to check any explicit formula for a recursive sequence is to plug in $n=1$ and $n=2$ and compare with the first two terms from the recursion. That simple test catches sign errors, off-by-one mistakes, and wrong starting values before you commit.
A sequence is defined by $a_n = 12 - 3(n+1)$ for $n \ge 0$ . Which recursive formula with $a_0$ as the initial value generates the same sequence?
Answer: $a_0=9, a_n=a_{n-1}-3$ for $n \ge 1$ Whenever you see a sequence defined by a linear explicit formula, recognize it as an arithmetic sequence: the number multiplied by $n$ is the common difference, and plugging in $n=0$ gives the initial value. Here $a_n = ́12 - ́3(n+1)$ . Simplify: $\,a_n = ́12 - ́3n - ́3 = ́9 - ́3n$ . So $a_0 = ́9$ , and each step decreases by $3$ , giving $a_0=9,\quad a_n=a_{n-1}-3 \text{ for } n\ge 1.$ The choice with $a_0=9,\ a_n=a_{n-1}-3$ is exactly the correct recursive form. The choice with $a_0=12,\ a_n=a_{n-1}-3$ uses the right common difference but wrong starting value; $12$ would come from ignoring the extra $-3$ in the formula. The choice with $a_0=9,\ a_n=a_{n-1}+3$ has the right initial value but wrong direction; adding $3$ yields an increasing sequence, whereas computing terms like $a_1=6,\ a_2=3$ shows values decrease. The choice with $a_0=6,\ a_n=a_{n-1}-3$ has the right common difference but initial value $6$ is what you might get by plugging in $n=1$ , not $n=0$ . Study tip: for an arithmetic sequence, get the first term and the common difference — the constant subtracted from one term to the next — then write the recursive rule as $a_0=\text{first},\, a_n=a_{n-1}+\text{difference}$ .
A sequence is defined by $a_1=4$ and $a_n=3a_{n-1}-2$ for $n\ge2$ . Which explicit formula is equivalent to this recursive definition?
Answer: $a_n=3^n+1$ The fixed point of the recurrence is 1 because $L=3L-2$ , so $L=1$ . Subtracting 1 from both sides gives $a_n-1=3(a_{n-1}-1)$ . Since $a_1-1=3$ , it follows that $a_n-1=3\cdot3^{n-1}=3^n$ , so $a_n=3^n+1$ . The formula $3^n-1$ shifts the constant the wrong way; $3^{n-1}+3$ matches only the first term; $2\cdot3^{n-1}+2$ treats the recurrence as purely geometric with an extra added constant.
An arithmetic sequence is defined by $a_1=-3$ and $a_n=a_{n-1}+d$ for $n\ge2$ . If $a_{10}=33$ , which explicit formula gives $a_{n+1}$ directly in terms of $n$ ?
Answer: $4n-3$ Since $a_{10}=a_1+9d$ , we have $33=-3+9d$ , so $d=4$ . Therefore $a_k=-3+4(k-1)=4k-7$ . Replacing $k$ with $n+1$ gives $a_{n+1}=4(n+1)-7=4n-3$ . The option $4n-7$ gives $a_n$ , $4n+1$ gives $a_{n+2}$ , and $3n-6$ comes from using the wrong common difference.
An explicit formula is $a_n=5\cdot3^{n-1}-2$ . If the same sequence is written in the recursive form $a_n=k\,a_{n-1}+c$ for $n\ge2$ , what are $k$ and $c$ ?
Answer: $k=3, c=4$ Substituting into $a_n=3a_{n-1}+4$ gives $3(5\cdot3^{n-2}-2)+4=5\cdot3^{n-1}-2$ , so $k=3$ and $c=4$ . The constant $c$ is not simply the explicit formula's constant term $-2$ , nor is it the coefficient $5$ from the explicit formula.
Which explicit formula can be represented by a recursive formula of the form $a_n=r\,a_{n-1}+d$ with constant values of $r$ and $d$ ?
Answer: $a_n=3(2^n)-1$ The formula $a_n=3(2^n)-1$ can be written as $a_n=2a_{n-1}+1$ because $2(3\cdot2^{n-1}-1)+1=3\cdot2^n-1$ . Polynomial formulas such as $2n^2+1$ and $n^3+1$ have nonconstant differences, and $2^n+n$ cannot be produced by a constant-coefficient first-order recurrence of this form.
An arithmetic sequence is defined by $t_1=8$ and $t_n=t_{n-1}-5$ for $n\ge2$ . Which explicit formula gives $t_{n+1}$ directly in terms of $n$ ?
Answer: $8-5n$ The explicit formula for the original index is $t_k=8-5(k-1)=13-5k$ . Replacing $k$ with $n+1$ gives $t_{n+1}=13-5(n+1)=8-5n$ . The expression $13-5n$ is $t_n$ , $3-5n$ is $t_{n+2}$ , and $8-5(n-1)$ is also $t_n$ .
The explicit formula $a_n=6n-4$ is defined for integers $n\ge0$ . Which recursive formula represents the same sequence over that domain?
Answer: $a_0=-4, a_n=a_{n-1}+6$ For $n=0$ , $a_0=-4$ , and each step increases by 6, so the recursive form is $a_0=-4, a_n=a_{n-1}+6$ . The option starting at $a_1=2$ generates the same tail but omits $a_0$ , so it does not represent the sequence over the stated domain. The options using 6 as a multiplier produce terms such as $-28$ or 44, which are not in the sequence.
The balance of an investment after $n$ years is modeled by $B_n = 500(1.02)^n$ for $n \ge 0$ . Which recursive formula with $B_0$ as the initial value gives the same balances?
Answer: $B_0=500, B_n=1.02B_{n-1}$ for $n \ge 1$ Whenever you see an explicit formula like $B_n = 500(1.02)^n$ , it describes exponential growth: the initial balance is $500$ , and each year the balance is multiplied by $1.02$ . So a matching recursive formula must start with $B_0=500$ and use the rule "multiply the previous balance by the same growth factor." That is exactly $B_n=1.02B_{n-1}$ for $n\ge 1$ . Why the others fail? The choice $B_n=1.02^nB_{n-1}$ uses a changing multiplier each step, so it grows too fast: after two years it gives $500 \cdot 1.02^3$ instead of $500 \cdot 1.02^2$ . The choice $B_n=B_{n-1}+0.02$ treats the 2% as an addition of $0.02$ dollars, producing a linear sequence like $500, 500.02, 500.04$ , not the exponential values $510, 520.2$ . The choice with $B_1=500$ and $B_n=1.02B_{n-1}$ has the right multiplier but the wrong starting year: since $B_1=1.02B_0$ , setting $B_1=500$ makes $B_0\approx 490.20$ , so it models a different initial investment. Only the formula with $B_0=500$ and multiplier $1.02$ reproduces the explicit balances exactly. Your takeaway: to convert an explicit exponential formula $B_n=a\cdot r^n$ into a recursive one, use $B_0=a$ and $B_n=rB_{n-1}$ . Watch for percent growth meaning "multiply by $1+$ percent," not "add the percent."
A student claims that the sequence $2,5,8,11,\ldots$ can be generated by the recursive formula $a_1=2, a_n=3a_{n-1}-1$ . Which statement correctly diagnoses the claim?
Answer: It is incorrect; the common difference is 3, so the formula should be $a_1=2, a_n=a_{n-1}+3$ . The sequence is arithmetic with common difference 3, so the correct recursive form is $a_1=2, a_n=a_{n-1}+3$ . The student's formula happens to give the second term, 5, but gives 14 for the third term instead of 8. The multiplier 3 and the constant 1 come from confusing the explicit formula's coefficient with a common ratio.
A sequence is defined by $a_1=1, a_2=5, a_n=3a_{n-1}-2a_{n-2}$ for $n\ge3$ . Which explicit formula is equivalent?
Answer: $a_n=2^{n+1}-3$ The characteristic equation is $r^2-3r+2=0$ , so $r=1$ and $r=2$ . Thus $a_n=A+B\cdot2^n$ . Using $a_1=1$ and $a_2=5$ gives $A=-3$ and $B=2$ , so $a_n=2^{n+1}-3$ . The other formulas fail the given first two terms or have the wrong sign on the constant.
A sequence is defined by $a_1=7$ and $a_{n+1}=a_n+3$ for $n \ge 1$ . Which explicit formula gives $a_n$ for $n \ge 1$ ?
Answer: $a_n=3n+4$ Whenever you see a sequence defined by a starting value plus a repeated addition, think arithmetic sequence. The key is identifying the first term $a_1$ and the common difference $d$ , then using the general form $a_n=a_1+(n-1)d$ . Here $a_1=7$ and each step adds $3$ , so $d=3$ . Substitute: $a_n=7+3(n-1)=7+3n-3=3n+4$ . This matches the correct choice $a_n=3n+4$ . A quick check: for $n=1$ , it gives $3(1)+4=7$ , exactly the first term. . Now the wrong formulas: $a_n=3n+10$ would give $a_1=13$ , so it seems to use $a_1+3(n+1)$ — an index shift too far. $a_n=3n+7$ would give $a_1=10$ , because it forgets to subtract $1$ from $n$ ; this is the common error of using $a_1+dn$ instead of $a_1+d(n-1)$ . $a_n=7n+3$ would give $a_1=10$ and increases by $7$ each time, confusing the first term with the common difference — those roles are not interchangeable. . A stronger approach: any time you need an explicit formula, evaluate it at $n=1$ and at $n=2$ . The correct formula must produce the givens first term and common difference. This quick check catches index shifts and reversed coefficients instantly.
Which recursive definition produces the same sequence as the explicit formula $b_n=5\cdot2^n-3$ for $n\ge1$ ?
Answer: $b_1=7, b_n=2b_{n-1}+3$ The first term is $b_1=5\cdot2-3=7$ . Substituting into $b_n=2b_{n-1}+3$ gives $2(5\cdot2^{n-1}-3)+3=5\cdot2^n-3$ , so it matches the explicit formula. The option using $-3$ changes the constant term; options with initial value 2 or 5 do not match $b_1=7$ .
Which recursive formula is equivalent to $a_n=4(-2)^n+1$ for $n\ge1$ ?
Answer: $a_1=-7, a_n=-2a_{n-1}+3$ The first term is $a_1=4(-2)+1=-7$ . Substituting into $a_n=-2a_{n-1}+3$ gives $-2(4(-2)^{n-1}+1)+3=4(-2)^n+1$ , so it matches. The options with constant $-1$ or $1$ produce the wrong second term, and the option with $a_1=-8$ has the wrong initial value.
A sequence is defined by $a_n = 5(2)^{n+1}$ for $n \ge 1$ . Which recursive formula generates the same sequence?
Answer: $a_1=20, a_n=2a_{n-1}$ for $n \ge 2$ Whenever you see a sequence given by an explicit formula like $a_n = 5(2)^{n+1}$ , your task is to translate it into recursive language: find the first term, then identify how each term is built from the previous one. The structure $a_n = 5 \cdot 2^{n+1}$ is exponential with base 2, so the common ratio is 2. To find the starting term, plug in $n=1$ : $a_1 = 5(2)^{1+1} = 5 \cdot 4 = 20$ . That means the sequence begins at 20 and every later term is twice the one before it, so the recursive form is $a_1 = 20,\ a_n = 2a_{n-1}$ for $n \ge 2$ . The wrong choices each come from a common slip. Starting with $a_1=10$ would be correct if the formula were $5(2)^n$ , but the extra $n+1$ exponent doubles the first term to 20. Starting with $a_1=40$ gives the value of $a_2$ , not $a_1$ , because $a_2 = 5(2)^3 = 40$ . Starting with $a_1=5$ ignores the power-of-2 factor entirely, treating the base coefficient as the first term. Each of those incorrect initial terms would produce a sequence shifted, doubled, or scaled incorrectly. Your quick strategy: given $a_n = a_1 \cdot r^{n-1}$ , identify $a_1$ by substituting $n=1$ , and recognize the base of the exponent as the recursive multiplier. Then write "first term, multiply by that ratio" — and check your first term before moving on.
A sequence is defined by $b_0 = 6$ and $b_n = -3b_{n-1}$ for $n \ge 1$ . Which explicit formula gives $b_n$ for $n \ge 0$ ?
Answer: $b_n=6(-3)^n$ Whenever you see a recursive rule like $b_n = -3b_{n-1}}$ , recognize it as a geometric sequence: the ratio is constant (-3), and the initial term is $b_0=6$ . For a geometric sequence, the explicit formula is $b_n = b_0 r^n$ , so $b_n=6(-3)^n$ . That formula works: at $n=0$ , it gives $6$ ; at $n=1$ , it gives $-18$ , exactly $-3\cdot 6$ . Now check the wrong options. The choice $6(-3)^{n-1}}$ uses an off-by-one exponent. It gives $n=0$ as $6(-3)^{-1}=-2$ , not $6$ , and it would only match if indices started at $1$ . The choice $(-18)^n$ multiplies the first term and ratio together as the base, giving $n=0$ as $1$ , not $6$ , and $n=2$ as $324$ , while the real value is $54$ . It confuses the first term $b_1=-18$ with the initial term. The choice $6(-3)^{n+1}}$ shifts the exponent in the opposite direction: at $n=0$ it gives $-18$ , so it actually represents the sequence starting with $b_1$ rather than $b_0$ . A quick habit: whenever you write or choose an explicit formula for a recursive sequence, test it at $n=0$ and $n=1$ . If it gives the stated initial term and the first recursive term, it is likely correct. Here, only $6(-3)^n$ passes both checks.
A sequence is defined by $a_n=\frac{3\cdot 2^{n+2}}{8}$ for $n \ge 1$ . Which recursive formula generates the sequence?
Answer: $a_1=3, a_n=2a_{n-1}$ for $n \ge 2$ Whenever you see a sequence defined by an explicit formula, your first move should be to simplify it and then find the first term and the common ratio between consecutive terms. Here, notice that $8=2^3$ , so $a_n=\frac{3\cdot 2^{n+2}}{2^3}=3\cdot 2^{n-1}.$ This is a geometric sequence: it starts at $a_1=3\cdot 2^{0}=3$ , and each next term is multiplied by $2$ . For example, $a_1=3$ , $a_2=6$ , $a_3=12$ , and so on. The recursive formula must therefore state $a_1=3$ and use the rule $a_n=2a_{n-1}$ for $n\ge2$ . The choice that starts with $a_1=12$ and doubles gives $12,24,48,\dots$ , which has the right ratio but the wrong starting point. The choice that starts with $a_1=6$ and doubles gives $6,12,24,\dots$ , again missing the true first term of $3$ . The choice that starts with $a_1=3$ but uses $a_n=\frac12 a_{n-1}$ produces $3,1.5,0.75,\dots$ , which reverses the growth into decay. Only the formula with $a_1=3$ and $a_n=2a_{n-1}$ matches both the first term and the doubling pattern. A quick strategy: to convert an explicit geometric formula to a recursive one, simplify the expression first, then compute $a_1$ and the ratio $a_n/a_{n-1}$ . That ratio becomes the multiplier in the recursive rule.

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