Question 1206279
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Recursive example:
{{{system(a[1] = 5,a[n+1] = a[n] + 7)}}}
The top line indicates the 1st term is 5.
The next line says "whatever the nth term is, add 7 to it to get the (n+1)<sup>th</sup> term".


{{{a[n]}}} = nth term
n is some positive integer 1,2,3,4,...
{{{a[n+1]}}} = term just after the nth term


So,
{{{a[2] = a[1]+7 = 5+7 = 12}}} is the 2nd term
{{{a[3] = a[2]+7 = 12+7 = 19}}} is the 3rd term
{{{a[4] = a[3]+7 = 19+7 = 26}}} is the 4th term
And so on.


We generate the sequence 5, 12, 19, 26, ...


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Since we're adding the same thing to each term, we can state this sequence is arithmetic.
{{{a[1] = 5}}} = first term
{{{d = 7}}} = common difference


The explicit closed form equation would be
{{{a[n] = a[1] + d(n-1)}}}
{{{a[n] = 5 + 7(n-1)}}}


Let's try n = 4
{{{a[n] = 5 + 7(n-1)}}}
{{{a[4] = 5 + 7(4-1)}}}
{{{a[4] = 5 + 7(3)}}}
{{{a[4] = 5 + 21}}}
{{{a[4] = 26}}}
That matches the previous {{{a[4]}}} value found through the recursive process.
I'll let the student try other values of n. Make sure of course to only plug in positive integers.


There's pros and cons to both forms. If you wanted say the 99th term of this arithmetic sequence, then it would be tedious to follow the recursive route. 
This is because you'd need the 98th term, but the 98th term needs the 97th term, and so on.
Use the explicit equation instead.


But the recursive method isn't always bad. It might be more intuitive to some students to see what is going on. For some students it might be easier to repeatedly add 7 to get to the desired term. The recursive form is generally practical for small values of n.
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