What's the relationship between linear,quadratic, arithmetic and geometric patterns?

Question

I've only studied linear and quadratic patterns. Now I'm studying sequences where it isn't defined whether the sequence is arithmetic, geometric, or quadratic. I'm given the formula $t=n^2 +n$. It looks similar to linear pattern's $y=mx + c$. Please help. Thanks

Answer 1

A sequence $a_1, a_2, \ldots$ can be viewed as a function $f$ on the positive integers. For example, if $a_n = n^2+n$, then $f(x)=x^2+x$ and $f$ is a quadratic function. A linear function has the form $f(x)=mx+b$, as you mention. Recall that in arithmetic and geometric sequences the difference or ratio between successive terms, respectively, is some fixed constant. If they don't mention what kind of sequence it is, it is probably just an arbitrary sequence $a_1, a_2, \ldots$.

Answer 2

The sequence $t=n^2+n$ is the sum to the $n$-th term of the arithmetic progression (AP) $2, 4, 6, 8, ..., 2n, ...$.

Proof:

For the AP given above, $$S=\frac n2(\ell+a)=\frac n2(2n+2)=n^2+n=t\qquad\blacksquare$$

Another Proof: $$t=n^2+n=2\left[\frac{n(n+1)}2\right]=2\sum_{r=1}^nr=\sum_{r=1}^n 2r\qquad\blacksquare$$

NB: - A formula with a quadratic in $n$ is the sum of an AP, as the sum can be written as $$S=\frac n2[2a+(n-1)d]=\frac n2 [(2a-d)+dn]=\left(a-\frac d2\right)n+\frac d2 n^2$$ In the case above, equating coefficients of $n$ and $n^2$ to $1$ gives $a=2, d=2$. This is the general approach but for the given example it is more laborious compared to the solutions above.