For the sequence 3,8,15,24,35....

find a formula for tsubn in terms of n.

3,8,15,24,35

1,2,3, 4, 5

I think it's

t(n) = (n+2) * n

To find a formula for the nth term of the sequence 3, 8, 15, 24, 35..., we need to look for a pattern in the numbers.

Let's examine the differences between consecutive terms:
8 - 3 = 5
15 - 8 = 7
24 - 15 = 9
35 - 24 = 11

From this observation, we can see that the differences are increasing by 2 each time. This suggests that the differences follow an arithmetic sequence with a common difference of 2.

To generate the differences, we can start with a common difference of 2 and add it successively:
2, 4, 6, 8, ...

Now let's add these differences to the previous term to get the next term:
3 + 2 = 5
8 + 4 = 12
15 + 6 = 21
24 + 8 = 32

We can see that this gives us the correct terms of the sequence:
3, 8, 15, 24, 35,...

So, the pattern we have identified shows that the nth term of the sequence can be found by adding the (n-1)th term to a term from an arithmetic sequence with a common difference of 2.

In formula notation, this can be expressed as:
t(sub)n = t(sub)n-1 + (2n - 2)

Therefore, the formula for the nth term (t(sub)n) of the sequence 3, 8, 15, 24, 35... is t(sub)n = t(sub)n-1 + (2n - 2).