Question #161960

Write the first 4 terms of the following formula, start with n = 1.

  1. an = 5n
  2. nbn = 3!bn = 3n bn = 3n^2 +2n - 6
  3. gn = 1 × 2 × ... n
  4. c1 = 2.5, cn = cn-1 + 1.5
  5. d1 = -3, dn = -2dn-1 + 1
1
Expert's answer
2021-02-11T12:44:50-0500

Given sequences are:-

1.an=5n1. a_n=5n\\

at n=1, a1=5a_1=5

Similarly a2=10,a3=15,a4=20a_2=10, a_3=15, a_4=20

First four terms are 5,10,15,20.5,10, 15, 20.


2.nbn=3!bn=3nbn=3n2+2n6nbn = 3!bn = 3n bn = 3n^2 +2n - 6

at n=1

First term is 3(1)2+2(1)6=13(1)^2+2(1)-6=-1

Similarly other terms at n=2,3,4n=2,3,4 are 10,27,5010,27,50

First four terms are 1,10,27,50-1,10,27,50


3.gn=1×2..×ng_n=1\times2..\times n

at n=1, g1=1g_1=1

Similarly at n=2,3,4

g2=2,g3=6,g4=24g_2=2,g_3=6,g_4=24


First four terms are 1,2,6,241,2,6,24


4.c1=2.5,cn=cn1+1.5c_1=2.5, c_n=c_{n-1}+1.5

at n=2, c2=c1+1.5=2.5+1.5=4c_2=c_1+1.5=2.5+1.5=4

at n=3, c3=c2+1.5=4+1.5=5.5c_3=c_2+1.5=4+1.5=5.5

at n=4, c4=c3+1.5=5.5+1.5=7c_4=c_3+1.5=5.5+1.5=7

First four terms are 2.5,4,5.5,72.5,4,5.5,7


5.d1=3,dn=2dn1+1d_1=-3,d_n=-2d_{n-1}+1

at n=2, d2=2d1+1=2(3)+1=7d_2=-2d_1+1=-2(-3)+1=7

at n=3, d3=2d2+1=2(7)+1=13d_3=-2d_2+1=-2(7)+1=-13

at n=4, d4=2d3+1=2(13)+1=27d_4=-2d_3+1=-2(-13)+1=27

First Four terms are 3,7,13,27-3, 7, -13,27


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