Question 1191041
Let the place count be {{{n}}}
Let the {{{n}}}th term be {{{a[n]}}}

Given {{{f(n=1)=6}}}

We are also told that any one term is the previous term + {{{2}}}.

This is derived from {{{f(n)=f(n-1)+2}}} where {{{f(n-1) }}}is the previous term.

Consequently we have an Arithmetic sequence with common difference of +{{{2}}}

From this the sequence is:

{{{n=1}}}→{{{a[1]=6}}}← given value

{{{n=2}}}→{{{a[2]=6+2=8}}}
{{{n=3}}}→{{{a[3]=6+2+2=10}}}
{{{highlight(n=4)}}}→{{{highlight(a[4]=6+2+2+2=12)}}}


so, the value of {{{highlight(f(4))}}} is {{{highlight(12)}}}


And so on. Also from this we also have an alternative equation for any an in that we have:

{{{a[n]=6+2(n-1)}}}