MATH122

Using the TI-83 to Evaluate Sequences

To evaluate any sequence, change the MODE setting from Func to Seq, then press the [Y =] key. As many as three sequences can be defined; they are named u, v, and w. Subscripts are put inside parenthesis. The value of *n*Min is the subscript value that begins the sequence, usually 0 or 1. Enter the formula for the sequence (see examples below), then go to the [TBLSET] screen. Enter values of TblStart = 1, DTbl = 1; set Indpnt: Auto and Depend: Auto. You may change the value of TblStart to any index value for which the sequence is defined if you want to look at terms of the sequence further along in the list.

Go to the [TABLE] screen. The values of the first few members of the sequence are given. More can be found by scrolling down.

A graph of values can be shown by using the [WINDOW] settings with *n*Min and *n*Max values that are consistent with the Xmin and Xmax values. It is up to the user to set the Ymin and Ymax values so the sequence values show on the screen just as it is when graphing other functions.

There are two type of sequences that can be evaluated on the TI-83: non-recursive and recursive. Here is an example for a non-recursive sequence.

Example 1: The sequence is {*a*_{n}} where *a*_{n} = 2 + 3n + (-1)^{n} for n ³ 1.
On the [Y=] screen, enter *n*Min = 1 and u(*n*) = 2 + 3n + (-1)^n (Use the same [X,T,q,*n*] key that you usually use to enter 'x' to enter the '*n*')

Recursively defined functions also use the [MODE] setting of Seq. Here are two examples for recursively-defined sequences.

Example 2: If a(1) = 5 and a(n+1) = 2+a(n) for n ³ 1, we use the original starting value for the subscript (*n*Min = 1), but we must write u(*n*) instead of a(n+1), so we subtract 1 from all subscript values in a to get the corresponding u information. a(n+1) = 2 +a(n) becomes u(*n*) = 2+u(*n*-1). (Use [2^{nd}][7] to enter 'u' in the formula, and use the same [X,T,q,*n*] key that you usually use to enter 'x' to enter the '*n*') Enter u(*n*Min) = 5; it is still true for u as it was for a(1) = 5.

Example 3: If a(1) = 5, a(2) = 12, and a(n+2) = a(n+1) +a(n) for n ³ 1, we use the original starting value of the subscript (*n*Min = 1), but we must write u(*n*) instead of a(n+2), so we subtract 2 from all subscript values in a to get the corresponding u information. a(n+2) = a(n+1) + a(n) becomes u(*n*) = u(*n*-1) + u(*n*-2). There are two initial values, so we use u(*n*Min) = {12,5} Note that the initial values for a, which are 5 and 12 respectively, are entered in reverse order for u!!