Question

Consider the sequence defined recursively by do = 0, an = an-1 + 3n – 1. a) Write out the first 5 terms of this sequence,

Answer 1

The first 5 terms of the sequence is 2,7,15,26,40.

Explanation:

Given : Consider the sequence defined recursively by
`a_0=0`
`a_n=a_(n-1)+3n-1`

To find : Write out the first 5 terms of this sequence ?

Solution :

`a_n=a_(n-1)+3n-1` and
`a_0=0`

The first five terms in the sequence is at n=1,2,3,4,5

For n=1,

`a_1=a_(1-1)+3(1)-1`

`a_1=a_(0)+3-1`

`a_1=0+2`

`a_1=2`

For n=2,

`a_2=a_(2-1)+3(2)-1`

`a_2=a_(1)+6-1`

`a_2=2+5`

`a_2=7`

For n=3,

`a_3=a_(3-1)+3(3)-1`

`a_3=a_(2)+9-1`

`a_3=7+8`

`a_3=15`

For n=4,

`a_4=a_(4-1)+3(4)-1`

`a_4=a_(3)+12-1`

`a_4=15+11`

`a_4=26`

For n=5,

`a_5=a_(5-1)+3(5)-1`

`a_5=a_(4)+15-1`

`a_5=26+14`

`a_5=40`

The first 5 terms of the sequence is 2,7,15,26,40.