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 \(a_{1}=7, \, a_{n+1}=a_{n}-4\)
\(a_n=11-4n\)
\(\,\,\,\,\,\text{Common difference } d=-4\)
\(\,\,\,\,\,a_n=a_1+d(n-1)=7-4(n-1)=11-4n\)
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