Recursive sequences define each term by using one or more previous terms. These problems practice finding terms from recursive rules, recognizing arithmetic and geometric recursive patterns, and rewriting some recursive rules as explicit formulas. Recursive sequences are useful for modeling patterns like Fibonacci numbers, repeated growth, alternating signs, factorials, and sequences that approach a limit.
Notes
Arithmetic Recursive Series
\(a_n=a_{n-1}+d\)
Geometric Recursive Series
\(a_n=r \cdot a_{n-1}\)
Practice Problems
\(\textbf{1)}\) Find the first 5 terms of this sequence \(a_{1}=2, \, a_{n+1}=a_{n}+3\)
The first 5 terms are \(2,5,8,11,14\)
\(\,\,\,\,\,a_{1}=2, \, a_{n+1}=a_{n}+3\)
\(\,\,\,\,\,a_{1}=2\)
\(\,\,\,\,\,a_2=a_1+3=(2)+3=5\)
\(\,\,\,\,\,a_3=a_2+3=(5)+3=8\)
\(\,\,\,\,\,a_4=a_3+3=(8)+3=11\)
\(\,\,\,\,\,a_5=a_4+3=(11)+3=14\)
The first 5 terms are \(2,5,8,11,14\)
\(\textbf{2)}\) Find the first 5 terms of this sequence \(a_{1}=-3, \, a_{n+1}=a_{n}+2n\)
The first 5 terms are \(-3,-1,3,9,17\)
\(\,\,\,\,\,a_{1}=-3\)
\(\,\,\,\,\,a_{2}=a_{1}+2(1)=-3+2=-1\)
\(\,\,\,\,\,a_{3}=a_{2}+2(2)=-1+4=3\)
\(\,\,\,\,\,a_{4}=a_{3}+2(3)=3+6=9\)
\(\,\,\,\,\,a_{5}=a_{4}+2(4)=9+8=17\)
The first 5 terms are \(-3,-1,3,9,17\)
\(\textbf{3)}\) Find the first 5 terms of this sequence \(a_{1}=0, \, a_{n+1}=5a_{n}-8\)
The first 5 terms are \(0,-8,-48,-248,-1248\)
\(\,\,\,\,\,a_{1}=0\)
\(\,\,\,\,\,a_{2}=5a_{1}-8=5(0)-8=-8\)
\(\,\,\,\,\,a_{3}=5a_{2}-8=5(-8)-8=-48\)
\(\,\,\,\,\,a_{4}=5a_{3}-8=5(-48)-8=-248\)
\(\,\,\,\,\,a_{5}=5a_{4}-8=5(-248)-8=-1248\)
The first 5 terms are \(0,-8,-48,-248,-1248\)
\(\textbf{4)}\) Find the first 5 terms of this sequence \(a_{1}=5, \, a_{2}=3, \, a_{n+2}=a_{n+1}-2a_{n}\)
The first 5 terms are \(5,3,-7,-13,1\)
\(\,\,\,\,\,a_{1}=5, \, a_{2}=3\)
\(\,\,\,\,\,a_{3}=a_{2}-2a_{1}=3-2(5)=-7\)
\(\,\,\,\,\,a_{4}=a_{3}-2a_{2}=-7-2(3)=-13\)
\(\,\,\,\,\,a_{5}=a_{4}-2a_{3}=-13-2(-7)=1\)
The first 5 terms are \(5,3,-7,-13,1\)
\(\textbf{5)}\) Write an explicit rule for the sequence \(a_{1}=3, \, a_{n}=4a_{n-1}\)
\(a_n=3(4)^{n-1}\)
\(\,\,\,\,\,\text{Geometric with ratio } r=4\)
\(\,\,\,\,\,a_n=a_1 r^{n-1}=3(4)^{n-1}\)
\(\textbf{6)}\) Find the first 8 terms of this sequence \(a_{1}=1, \, a_{2}=1, \, a_{n+2}=a_{n+1}+a_{n}\)
The first 8 terms are \(1,1,2,3,5,8,13,21\)
\(\,\,\,\,\,a_{1}=1, \, a_{2}=1\)
\(\,\,\,\,\,a_{3}=a_{2}+a_{1}=1+1=2\)
\(\,\,\,\,\,a_{4}=a_{3}+a_{2}=2+1=3\)
\(\,\,\,\,\,a_{5}=a_{4}+a_{3}=3+2=5\)
\(\,\,\,\,\,a_{6}=a_{5}+a_{4}=5+3=8\)
\(\,\,\,\,\,a_{7}=a_{6}+a_{5}=8+5=13\)
\(\,\,\,\,\,a_{8}=a_{7}+a_{6}=13+8=21\)
The first 8 terms are \(1,1,2,3,5,8,13,21\)
\(\textbf{7)}\) Find the first 6 terms of this sequence \(a_{1}=2, \, a_{n+1}=(-1)^{n}a_{n}+1\)
The first 6 terms are \(2,-1,0,1,2,-1\)
\(\,\,\,\,\,a_{1}=2\)
\(\,\,\,\,\,a_{2}=(-1)^{1}(2)+1=-2+1=-1\)
\(\,\,\,\,\,a_{3}=(-1)^{2}(-1)+1=1-1=0\)
\(\,\,\,\,\,a_{4}=(-1)^{3}(0)+1=0+1=1\)
\(\,\,\,\,\,a_{5}=(-1)^{4}(1)+1=1+1=2\)
\(\,\,\,\,\,a_{6}=(-1)^{5}(2)+1=-2+1=-1\)
The first 6 terms are \(2,-1,0,1,2,-1\)
\(\textbf{8)}\) Find the first 6 terms of this sequence \(a_{1}=1, \, a_{n+1}=a_{n}+(2n+1)\)
The first 6 terms are \(1,4,9,16,25,36\)
\(\,\,\,\,\,a_{1}=1\)
\(\,\,\,\,\,a_{2}=1+(2\cdot1+1)=1+3=4\)
\(\,\,\,\,\,a_{3}=4+(2\cdot2+1)=4+5=9\)
\(\,\,\,\,\,a_{4}=9+(2\cdot3+1)=9+7=16\)
\(\,\,\,\,\,a_{5}=16+(2\cdot4+1)=16+9=25\)
\(\,\,\,\,\,a_{6}=25+(2\cdot5+1)=25+11=36\)
The first 6 terms are \(1,4,9,16,25,36\)
\(\textbf{9)}\) Find the first 6 terms of this sequence \(a_{1}=6, \, a_{n+1}=-2a_{n}\)
The first 6 terms are \(6,-12,24,-48,96,-192\)
\(\,\,\,\,\,a_{1}=6\)
\(\,\,\,\,\,a_{2}=-2(6)=-12\)
\(\,\,\,\,\,a_{3}=-2(-12)=24\)
\(\,\,\,\,\,a_{4}=-2(24)=-48\)
\(\,\,\,\,\,a_{5}=-2(-48)=96\)
\(\,\,\,\,\,a_{6}=-2(96)=-192\)
The first 6 terms are \(6,-12,24,-48,96,-192\)
\(\textbf{10)}\) Find the first 5 terms of this sequence \(a_{1}=1, \, a_{n+1}=a_{n}+\dfrac{1}{n}\)
The first 5 terms are \(1,2,\dfrac{5}{2},\dfrac{17}{6},\dfrac{37}{12}\)
\(\,\,\,\,\,a_{1}=1\)
\(\,\,\,\,\,a_{2}=1+\frac{1}{1}=2\)
\(\,\,\,\,\,a_{3}=2+\frac{1}{2}=\frac{5}{2}\)
\(\,\,\,\,\,a_{4}=\frac{5}{2}+\frac{1}{3}=\frac{15+2}{6}=\frac{17}{6}\)
\(\,\,\,\,\,a_{5}=\frac{17}{6}+\frac{1}{4}=\frac{34+3}{12}=\frac{37}{12}\)
The first 5 terms are \(1,2,\frac{5}{2},\frac{17}{6},\frac{37}{12}\)
\(\textbf{11)}\) Find the first 6 terms of this sequence \(a_{1}=2, \, a_{2}=4, \, a_{n+2}=3a_{n+1}-2a_{n}\)
The first 6 terms are \(2,4,8,16,32,64\)
\(\,\,\,\,\,a_{1}=2, \, a_{2}=4\)
\(\,\,\,\,\,a_{3}=3(4)-2(2)=12-4=8\)
\(\,\,\,\,\,a_{4}=3(8)-2(4)=24-8=16\)
\(\,\,\,\,\,a_{5}=3(16)-2(8)=48-16=32\)
\(\,\,\,\,\,a_{6}=3(32)-2(16)=96-32=64\)
The first 6 terms are \(2,4,8,16,32,64\)
\(\textbf{12)}\) Find the first 5 terms of this sequence \(a_{1}=1, \, a_{n+1}=3a_{n}+2\)
The first 5 terms are \(1,5,17,53,161\)
\(\,\,\,\,\,a_{1}=1\)
\(\,\,\,\,\,a_{2}=3(1)+2=5\)
\(\,\,\,\,\,a_{3}=3(5)+2=17\)
\(\,\,\,\,\,a_{4}=3(17)+2=53\)
\(\,\,\,\,\,a_{5}=3(53)+2=161\)
The first 5 terms are \(1,5,17,53,161\)
\(\textbf{13)}\) Find the first 6 terms of this sequence (factorial) \(a_{1}=1, \, a_{n+1}=n \cdot a_{n}\)
The first 6 terms are \(1,1,2,6,24,120\)
\(\,\,\,\,\,a_{1}=1\)
\(\,\,\,\,\,a_{2}=1\cdot1=1\)
\(\,\,\,\,\,a_{3}=2\cdot1=2\)
\(\,\,\,\,\,a_{4}=3\cdot2=6\)
\(\,\,\,\,\,a_{5}=4\cdot6=24\)
\(\,\,\,\,\,a_{6}=5\cdot24=120\)
The first 6 terms are \(1,1,2,6,24,120\)
\(\textbf{14)}\) Find the first 6 terms of this sequence (averaging toward a limit) \(a_{1}=10, \, a_{n+1}=\dfrac{a_{n}+4}{2}\)
The first 6 terms are \(10,7,\dfrac{11}{2},\dfrac{19}{4},\dfrac{35}{8},\dfrac{67}{16}\)
\(\,\,\,\,\,a_{1}=10\)
\(\,\,\,\,\,a_{2}=\dfrac{10+4}{2}=7\)
\(\,\,\,\,\,a_{3}=\dfrac{7+4}{2}=\dfrac{11}{2}\)
\(\,\,\,\,\,a_{4}=\dfrac{\frac{11}{2}+4}{2}=\dfrac{19}{4}\)
\(\,\,\,\,\,a_{5}=\dfrac{\frac{19}{4}+4}{2}=\dfrac{35}{8}\)
\(\,\,\,\,\,a_{6}=\dfrac{\frac{35}{8}+4}{2}=\dfrac{67}{16}\)
The first 6 terms are \(10,7,\frac{11}{2},\frac{19}{4},\frac{35}{8},\frac{67}{16}\)
\(\textbf{15)}\) Write an explicit rule for this arithmetic sequence \(a_{1}=7, \, a_{n+1}=a_{n}-4\)
\(a_n=11-4n\)
\(\,\,\,\,\,\text{Common difference } d=-4\)
\(\,\,\,\,\,a_n=a_1+d(n-1)=7-4(n-1)=11-4n\)
\(\textbf{16)}\) Write an explicit rule for this geometric sequence \(a_{1}=5, \, a_{n+1}=2a_n\)
\(a_n=5(2)^{n-1}\)
\(\,\,\,\,\,\text{Geometric with first term } a_1=5 \text{ and ratio } r=2\)
\(\,\,\,\,\,a_n=a_1r^{n-1}\)
\(\,\,\,\,\,a_n=5(2)^{n-1}\)
\(\textbf{17)}\) Find the first 5 terms of this sequence \(a_{1}=4, \, a_{n+1}=a_n^2-3\)
The first 5 terms are \(4,13,166,27553,759167806\)
\(\,\,\,\,\,a_{1}=4\)
\(\,\,\,\,\,a_{2}=4^2-3=16-3=13\)
\(\,\,\,\,\,a_{3}=13^2-3=169-3=166\)
\(\,\,\,\,\,a_{4}=166^2-3=27556-3=27553\)
\(\,\,\,\,\,a_{5}=27553^2-3=759167809-3=759167806\)
The first 5 terms are \(4,13,166,27553,759167806\)
\(\textbf{18)}\) Find the first 6 terms of this sequence \(a_{1}=3, \, a_{n+1}=a_n+n^2\)
The first 6 terms are \(3,4,8,17,33,58\)
\(\,\,\,\,\,a_{1}=3\)
\(\,\,\,\,\,a_{2}=a_1+1^2=3+1=4\)
\(\,\,\,\,\,a_{3}=a_2+2^2=4+4=8\)
\(\,\,\,\,\,a_{4}=a_3+3^2=8+9=17\)
\(\,\,\,\,\,a_{5}=a_4+4^2=17+16=33\)
\(\,\,\,\,\,a_{6}=a_5+5^2=33+25=58\)
The first 6 terms are \(3,4,8,17,33,58\)
\(\textbf{19)}\) Find the first 6 terms of this sequence \(a_{1}=1, \, a_{2}=2, \, a_{n+2}=2a_{n+1}+a_n\)
The first 6 terms are \(1,2,5,12,29,70\)
\(\,\,\,\,\,a_{1}=1, \, a_{2}=2\)
\(\,\,\,\,\,a_{3}=2a_2+a_1=2(2)+1=5\)
\(\,\,\,\,\,a_{4}=2a_3+a_2=2(5)+2=12\)
\(\,\,\,\,\,a_{5}=2a_4+a_3=2(12)+5=29\)
\(\,\,\,\,\,a_{6}=2a_5+a_4=2(29)+12=70\)
The first 6 terms are \(1,2,5,12,29,70\)
\(\textbf{20)}\) Find the first 5 terms of this sequence \(a_{1}=12, \, a_{n+1}=\dfrac{1}{2}a_n+3\)
The first 5 terms are \(12,9,\dfrac{15}{2},\dfrac{27}{4},\dfrac{51}{8}\)
\(\,\,\,\,\,a_{1}=12\)
\(\,\,\,\,\,a_{2}=\dfrac{1}{2}(12)+3=6+3=9\)
\(\,\,\,\,\,a_{3}=\dfrac{1}{2}(9)+3=\dfrac{9}{2}+\dfrac{6}{2}=\dfrac{15}{2}\)
\(\,\,\,\,\,a_{4}=\dfrac{1}{2}\left(\dfrac{15}{2}\right)+3=\dfrac{15}{4}+\dfrac{12}{4}=\dfrac{27}{4}\)
\(\,\,\,\,\,a_{5}=\dfrac{1}{2}\left(\dfrac{27}{4}\right)+3=\dfrac{27}{8}+\dfrac{24}{8}=\dfrac{51}{8}\)
The first 5 terms are \(12,9,\dfrac{15}{2},\dfrac{27}{4},\dfrac{51}{8}\)
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