Find the first five terms of the sequence for which a1=5,and an+1=3an+1.
an+1=3an+1
a1+1=3a1+1
a2=3(5)+1
a2=16
a2+1=3a2+1
a3=3(16)+1
a3=49
a3+1=3a3+1
a4=3(49)+1
a4=148
a4+1=3a4+1
a5=3(148)+1
a5=445
so the first five terms are: 5,16,49,148,445
Your answer is correct, but I am having difficulty following your notation.
How about instead of an+1=3an+1 something like
term(n+1) = 3term(n) + 1
The first five terms of the sequence are 5, 16, 49, 148, 445.
To find the first five terms of the sequence, we start with the given initial term, a1 = 5. Then, we can use the given recursive formula, an+1 = 3an + 1, to find the subsequent terms.
To find a2, we substitute n = 1 into the formula:
a2 = 3a1 + 1 = 3(5) + 1 = 16
To find a3, we substitute n = 2 into the formula:
a3 = 3a2 + 1 = 3(16) + 1 = 49
To find a4, we substitute n = 3 into the formula:
a4 = 3a3 + 1 = 3(49) + 1 = 148
To find a5, we substitute n = 4 into the formula:
a5 = 3a4 + 1 = 3(148) + 1 = 445
Therefore, the first five terms of the sequence are 5, 16, 49, 148, and 445.