Question:
Recursive definitions of sequence $a_n = n(n+1)$ and $a_n = n^2$.
My Attempt:
For the first one,$a_n = n(n+1)$, I first manually generate a sequence using $n \geq 1$,
$$2, 6, 12, 20, 30, ...$$
What I noticed is, that $n^{th}$ term is $a_{n-1} + 2n+ 4$.
For the 2nd, $a_n = n^2$, for $n\geq 1$. Since $n^2 = (n^2 + n) - n$, thus,
$$a_n = a_{n-1} + 2n + 4 - 1$$ $$a_n = a_{n-1} + 2n + 3$$
Since the -1 will accumulate to $n$.
Problems:
For both these two sequences, I have to go to the trouble of laying out sequences, and use of the sequences to build the other. It seems too much work for such simple equations, do you guys have a better and more straightforward solution to this?
Also if my solutions are wrong, pls point it out.
$\endgroup$ 14 Answers
$\begingroup$For the first: we have $$a_{n-1}=(n-1)n$$ so $$a_n-a_{n-1}=n(n+1)-n(n-1)=2n$$ so the sequence is defined by $$a_n=a_{n-1}+2n\quad;\quad a_1=2$$ Can you do the same method for the second sequence?
$\endgroup$ 2 $\begingroup$Your approach is to find something so that
$$a_n = a_{n-1} + \text{something}$$
so an easy method is to simply solve this equation.
$\endgroup$ $\begingroup$For the sequence $a_n = n(n+1)$, we have:
$a_n - a_{n-1} = n(n+1) - (n-1)n = (n^2+n)-(n^2-n) = 2n$
So, one possible recursive definition is "$a_n = a_{n-1} + 2n$ for all $n \ge 2$, $a_1 = 2$".
We can also compute the second difference:
$(a_n - a_{n-1}) - (a_{n-1} - a_{n-2}) = 2n - 2(n-1) = 2$
So, another recursive definition is "$a_n = 2a_{n-1} - a_{n-2} + 2$ for all $n \ge 3$, $a_1 = 2$, $a_2 = 6$".
Obviously, there are infinitely many ways to wire a recursive definition for this sequence. You should pick a way that works for your application.
You can do similar things with the sequence $b_n = n^2$.
$\endgroup$ $\begingroup$Hint:
- For the latter, try expressing $a_{n-1}\cdot a_{n+1}$ in terms of $a_n$
- For the former, try getting rid of the n by expressing it in terms of $\Big\lfloor\sqrt{a_{n-1}}\Big\rfloor$ or $\Big\lceil\sqrt{a_{n-1}}\Big\rceil$.
