Hello sean.1986...
Here is a link to LaTeX tutorial http://www.mathhelpforum.com/math-he...-tutorial.html
here n = 1,2,3...what do you mean by "are you allowed to include n in the recursive definition"?
The definition I gave for d will compute the square of n, but I got that but looking at the definition of a square number on wiki.
"The nth square number can be calculated from the previous two by doubling the (n − 1)-th square, subtracting the (n − 2)-th square number, and adding 2, because n^2 = 2(n − 1)^2 − (n − 2)^2 + 2. For example, 2×52 − 42 + 2 = 2×25 − 16 + 2 = 50 − 16 + 2 = 36 = 62."
Moo did the first one, I will try the remaining. Hopefully they match with sean's answers.
b)$\displaystyle a_{n+1} = 1+(-1)^{n+1} = 1-(-1)^n \Rightarrow a_n + a_{n+1} = 2 \Rightarrow a_{n+1} = 2 - a_n , a_1 = 0$
c)$\displaystyle a_{n-1} = n(n-1) , a_{n+1} = n(n+1) \Rightarrow a_n = a_{n-1} + 2n , a_1 = 2$
d) $\displaystyle a_{n+1} = n^2 + 2n + 1 \Rightarrow a_{n+1} = a_n + 2n + 1 , a_1 = 1$
A recursive definition of a sequence $\displaystyle \{a_n, n \in \mathbb{N} \}$ is a rule that tells you the $\displaystyle n$ th term in terms of one or more proceeding terms and possibly n, and a set of initial condition sufficient to allow you to apply the recursion to calculate the subsequent terms. Such a definition is not necessarilly unique.
Lets look at (a)
$\displaystyle a_n = 4n-2$
now $\displaystyle a_{n-1}=4n-6$, so:
$\displaystyle a_n=a_{n-1}+4$
and the initial condition we need to start is that $\displaystyle a_1=2$.
But we could also use two preceeding terms like:
$\displaystyle a_n=\frac{(a_{n-1}+a_{n-2})}{2}+6$
and the initial conditions we need to start is that $\displaystyle a_1=2$, and $\displaystyle a_2=6 $.
RonL