For a given pattern (1,4,9,16..)
What is the value for the nth number in the series and what is the pattern?
We have a difference in opinion with my son's 5th grade math teacher and want to get consensus.
$\endgroup$ 34 Answers
$\begingroup$This is the following succession: $$a_n=n^2$$.
$\endgroup$ $\begingroup$$$a_1=1$$
$$a_n=a_{n-1}+(2n-1)$$
is a recursive way of representing the sequence of squares.
$\endgroup$ 4 $\begingroup$as we can see, these are the square numbers of 1,2,3,4 and so on. so the first number is 1^2=1 2^2=4 3^2=9 4^2=16 and so on. for any nth term,the result is the square of it, so the pattern is n^2.
$\endgroup$ 0 $\begingroup$The following is overkill for this sequence of perfect squares, but in general...
Every sequence with a “constant acceleration” (second common difference) $d$ has a recurrence equation $$T_n-T_{n-1}=dn+s$$ for some $s.$
- So, its closed-form expression for $T_n$ can be derived usingtelescoping cancellation and summing up the ensuing arithmetic sequence.
Expressing $d$ and $s$ in terms of $T_1, T_2$ and $T_3$ gives the unwieldy $$T_n=\left(\frac{T_1-2T_2+T_3}2\right)n^2+\left(\frac{-5T_1+8T_2-3T_3}2\right)n+\left(3T_1-3T_2+T_3\right)\\=\frac d2\,n^2+\left(\frac{-5T_1+8T_2-3T_3}2\right)n+\left(3T_1-3T_2+T_3\right).$$
Thus, such sequences are commonly called quadratic sequences.
- So, another method to derive the closed-form expression for $T_n$ is to plug $n=1,2,3$ into $$T_n=pn^2+qn+r$$ and solve simultaneously for $p,q$ and $r,$ or, to plug $n=1,2$ into$$T_n=\frac d2n^2+qn+r$$ and solve simultaneously for $q$ and $r.$
