# Write the first six terms of the sequence given by : a1=a2=1 an=2*an-1 + 3*an-2, n>2 n n.n-1,n-2  is an index.

crmhaske | College Teacher | (Level 3) Associate Educator

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As the first two terms are known, we only need to find the next four terms as follows:

a1 = 1
a2 = 1
a3 = 2(a2) + 3(a1) = 2(1) + 3(1) = 5
a4 = 2(a3) + 3(a2) = 2(5) + 3(1) = 10 + 3 = 13
a5 = 2(a4) + 3(a3) = 2(13) + 3(5) = 26 + 15 = 41
a6 = 2(a5) + 3(a4) = 2(41) + 3(13) = 82 + 39 = 121

Therefore the first six terms are: 1, 1, 5, 13, 41, 121

hala718 | High School Teacher | (Level 1) Educator Emeritus

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a1= a2 = 1

an = 2*an-1 + 3*an-2

a1= 1

a2 = 1

a3= 2*a2 + 3*a1= 2*1 + 3*1 = 5

a4 = 2*a3 + 3*a2 = 2*5 + 3*1 = 13

a5 = 2*a4 + 3*a3 = 2*13 + 3*5 = 26+15 = 41

a6= 2*a5 + 3*a4 = 2*41 + 3*13 = 82+ 39 = 121

Then the first  6 terms are:

1, 1, 5,13, 41, 121

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Because the first and the second term are known, we'll have to write the next 4 terms.

We'll use the formula of the general term, given by enunciation:

an=2*an-1 + 3*an-2

Now, we'll put n = 3:

a3 = 2*a2 + 3*a1

a3 = 2*1 + 3*1

a3 = 2+3

a3 = 5

a4 = 2*a3 + 3*a2

a4 = 2*5 + 3*1

a4 = 10 + 3

a4 = 13

a5 = 2*a4 + 3*a3

a5 = 2*13 + 3*5

a5 = 26 + 15

a5 = 41

a6 = 2*a5 + 3*a4

a6 = 2*41 + 3*13

a6 = 82 + 39

a6 = 121

The 6 terms of the sequence are: 1, 1, 5, 13, 41, 121.

neela | High School Teacher | (Level 3) Valedictorian

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a1 = a2 =1

an = 2an-1 +3an-2  n > 2.

Applying the relation, for n =3,4,5,6

a3 = 2a2+3a1 =2+3 = 5,

a4 =2a3+3a2 = 2*5 +3*1 = 13

a5 = 2a4+3a3 = 2*13+3*5 =41

a6 = 2a5+3a4 = 2*41+3*13 = 121.

So 1,   1,   4,    13,   41, and 121  are the first 6 terms.