Question 1104290
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<pre>
1.  You are just given  that  {{{a[1]}}} = 1.


2.  Use the given formula to calculate  {{{a[2]}}}.

    You have  n= 2,  n-1= (2-1) = 1;  therefore, the formula takes the form

    {{{a[2]}}} = {{{1/(1+a[1])}}} = {{{1/(1+1)}}} = {{{1/2}}}.    (I substituted {{{a[1]}}} = 1).



3.  Now, when you know {{{a[2]}}}, you can calculate  {{{a[3]}}}, using the given formula.

    You have  n= 3,  n-1= (3-1) = 2;  therefore, the formula takes the form

    {{{a[3]}}} = {{{1/(1+a[2])}}} = {{{1/(1+(1/2))}}} = {{{1/(2/2+1/2))}}} = {{{1/((3/2))}}} = {{{2/3}}}.    (I substituted {{{a[2]}}} = {{{1/2}}}).



4.  Now, when you know {{{a[3]}}}, you can calculate  {{{a[4]}}}, using the given formula.

    You have  n= 4,  n-1= (4-1) = 3;  . . . and you can continue from this point on your own,  using the same logic . . . 
</pre>


It is how the recursive formulas work: if you know all preceding terms, you can calculate the next term, using the given recursive formula.



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