Finding a term in this series

[SimonSpacz]

The problem: Given this sequence Tn-1 = (Tn)^2 – 3n , given that T1=1, find T5

Tn referring to the nth term.
My work so far; Manipulated the formula of Tn=a+(n-1)d for T4 and T5 and filled them in to try and work out d (figuring that a=1, from t1=1)
I got decimal place answers for d that didnt work for the formula given in the question so I'm kinda lost and realize my lack of knowledge in this sector.
Any help?

 

[lookagain]

The problem: Given this sequence > > Tn-1 = (Tn)^2 – 3n < < , given that T1=1, find T5

Tn referring to the nth term.
My work so far; Manipulated the formula of

> > Tn=a+(n-1)d < <

for T4 and T5 and filled them in to try and work out d
(figuring that a=1, from t1=1) I got decimal place answers for d that didn't work for the formula given in the

question so I'm kinda lost and realize my lack of knowledge in this sector.
Any help?

If your first equation is

$\displaystyle T_{n - 1} \ = \ (T_n)^2 \ - \ 3n,$

then the first few terms show that the formula in the second
equation can't be used. The first few terms in the recursive equation do not have a common difference.
The common differences show up in arithmetic sequences. The second equation is used for
arithmetic sequences.

 

[daon2]

Is it possible you have the indices confused?

For example, $\displaystyle T_2 = \sqrt{7}$ and $\displaystyle T_2 = -\sqrt{7}$ both are acceptable answers.

 

[DrPhil]

The problem: Given this sequence Tn-1 = (Tn)^2 – 3n , given that T1=1, find T5

Tn referring to the nth term.
My work so far; Manipulated the formula of Tn=a+(n-1)d for T4 and T5 and filled them in to try and work out d (figuring that a=1, from t1=1)
I got decimal place answers for d that didnt work for the formula given in the question so I'm kinda lost and realize my lack of knowledge in this sector.
Any help?

Because of the square (or square root if you look the other direction) in the recursion formula, this is NOT an arithmetic sequence with equal spacing "d" between terms, That is, the terms don't have the form Tn=a+(n-1)d (as you discovered). Please do check the problem to be sure that what you type really looks like what was given. If you have it right, solving for T_n would give

$\displaystyle T_n = \sqrt{T_{n-1} + 3n}$

As daon2 pointed out, this is ambiguous unless there is something that tells you always to use the positive square root. If you made that assumption, you could in principle write down the sequence up through $\displaystyle T_5$. But this doesn't look "right" - very messy with the square roots.