The sequence 1, 4, 9, 16,...

Question

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.

Answer 1

This is the following succession: $$a_n=n^2$$.

Answer 2

$$a_1=1$$

$$a_n=a_{n-1}+(2n-1)$$

is a recursive way of representing the sequence of squares.

Answer 3

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.

Answer 4

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 using telescoping 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 $r.$