General Term for a Sequence

Hello, MATNTRNG!

I hope the $S$ is a typo . . .

Quote:

$\text{Find }a_n\text{ of }\{a_n\}.\;\text{ given: }\:a_1 = 1,\;\;a_{n+1} = a_n + 2n + 1$

Crank out the first few terms: . $1,\;6,\;13,\;22,\;33,\;46,\;61,\;\hdots$

Take the difference of consecutive terms,
. . then the differences of the differences, and so on.

. . $\begin{array}{cccccccccccc} \text{Sequence:} & 1 && 6 && 13 && 22 && 33 && 46 \\ \text{1stn diff:} && 5 && 7 && 9 && 11 && 13 \\ \text{2nd diff:} &&& 2 && 2 && 2 && 2 && 2 \end{array}$

Since the second differences are constant,
. . the generating function is of the second degree, a quadratic.

The general quadratic function is: . $f(n) \:=\:an^2 + bn + c$

Use the first three values of the function and substitute:

. . $\begin{array}{ccccc} f(3) = 13: & 9a + 3b + c &=& 13 & [1] \\ f(2) = 6: & 4a + 2b + c &=& 6 & [2] \\ f(1) = 1: & a + b + c & = & 1 & [3] \end{array}$

. . $\begin{array}{ccccc} \text{Subtract [1] - [2]:} & 5a + b &=& 7 & [4] \\ \text{Subtract [2] - [3]:} & 3a + b &=& 5 & [5] \end{array}$

. . $\text{Subtract [4] - [5]:} \;\;2a \:=\:2 \quad\Rightarrow\quad \boxed{a \:=\:1}$

Substitute into [5]: . $3(1) + b \:=\:5 \quad\Rightarrow\quad\boxed{ b \:=\:2}$

Substitute into [3]: . $1 + 2 + c \:=\:1 \quad\Rightarrow\quad\boxed{ c \:=\:-2}$

Therefore, the generating function is: . $f(n) \;=\;n^2 + 2n - 2$