Right now in my class i am learning about sequences and series and im currtently suck on this problem.
"A certain sequence is defined recursively by the t1=1 , t2=2, t2n=2t2n-2, t2n+1=3t2n-1. Find the first eight terms of the sequence."
thanks for your help.
Write the sixth term of the sequence: t1 = 1 and tn = 3tn-1 + 1
To solve this problem, we need to find the first eight terms of the given sequence.
Let's start by understanding the recursive definitions of the sequence:
1. t1 = 1
2. t2 = 2
3. tn = 2tn-2 (for even n, i.e., t2n)
4. tn = 3tn-1 (for odd n, i.e., t2n+1)
Using these definitions, we can proceed step by step to find the first eight terms:
1. t1 = 1 (given in the problem)
2. t2 = 2 (given in the problem)
3. t3 = 3t2-1 (using definition 4)
= 3t2 (since t2-1 equals t2 using the given definitions)
= 3(2) = 6
4. t4 = 2t4-2 (using definition 3)
= 2t2 (since t4-2 equals t2 using the given definitions)
= 2(2) = 4
5. t5 = 3t4-1 (using definition 4)
= 3t4 (since t4-1 equals t4 using the given definitions)
= 3(4) = 12
6. t6 = 2t6-2 (using definition 3)
= 2t4 (since t6-2 equals t4 using the given definitions)
= 2(4) = 8
7. t7 = 3t6-1 (using definition 4)
= 3t6 (since t6-1 equals t6 using the given definitions)
= 3(8) = 24
8. t8 = 2t8-2 (using definition 3)
= 2t6 (since t8-2 equals t6 using the given definitions)
= 2(8) = 16
So, the first eight terms of the sequence are: 1, 2, 6, 4, 12, 8, 24, 16.
I hope this explanation helps you understand how to solve the problem.