2021-10-16T13:10:37-04:00
Continue the two sequences of numbers below and find an equation to each of the sequences:
n 1 2 3 4 5 6 7 Equation
an 2 5 9 14 20 27
bn 1 3 12 60 360 2520
1
2021-10-19T15:59:02-0400
a n = a n − 1 + n + 1 , n ≥ 2 , a 1 = 1 a_n=a_{n-1}+n+1, n\geq2, a_1=1 a n = a n − 1 + n + 1 , n ≥ 2 , a 1 = 1
a 2 = a 1 + 2 + 1 = 2 + 3 = 5 a_2=a_1+2+1=2+3=5 a 2 = a 1 + 2 + 1 = 2 + 3 = 5
a 3 = a 2 + 3 + 1 = 5 + 4 = 9 a_3=a_2+3+1=5+4=9 a 3 = a 2 + 3 + 1 = 5 + 4 = 9
a 4 = a 3 + 4 + 1 = 9 + 5 = 14 a_4=a_3+4+1=9+5=14 a 4 = a 3 + 4 + 1 = 9 + 5 = 14
a 5 = a 4 + 5 + 1 = 14 + 6 = 20 a_5=a_4+5+1=14+6=20 a 5 = a 4 + 5 + 1 = 14 + 6 = 20
a 6 = a 5 + 6 + 1 = 20 + 7 = 27 a_6=a_5+6+1=20+7=27 a 6 = a 5 + 6 + 1 = 20 + 7 = 27
a 7 = a 6 + 7 + 1 = 5 + 4 = 27 + 8 = 35 a_7=a_6+7+1=5+4=27+8=35 a 7 = a 6 + 7 + 1 = 5 + 4 = 27 + 8 = 35
b n = b n − 1 ( n + 1 ) , n ≥ 2 , b 1 = 1 b_n=b_{n-1}(n+1), n\geq2, b_1=1 b n = b n − 1 ( n + 1 ) , n ≥ 2 , b 1 = 1
b 2 = b 1 ( 2 + 1 ) = 3 b_2=b_1(2+1)=3 b 2 = b 1 ( 2 + 1 ) = 3
b 3 = b 2 ( 3 + 1 ) = 12 b_3=b_2(3+1)=12 b 3 = b 2 ( 3 + 1 ) = 12
b 4 = b 3 ( 4 + 1 ) = 60 b_4=b_3(4+1)=60 b 4 = b 3 ( 4 + 1 ) = 60
b 5 = b 4 ( 5 + 1 ) = 360 b_5=b_4(5+1)=360 b 5 = b 4 ( 5 + 1 ) = 360
b 6 = b 5 ( 6 + 1 ) = 2520 b_6=b_5(6+1)=2520 b 6 = b 5 ( 6 + 1 ) = 2520
b 7 = b 6 ( 7 + 1 ) = 20160 b_7=b_6(7+1)=20160 b 7 = b 6 ( 7 + 1 ) = 20160
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