Question 1005810
If {{{a[n+1]=-4a[n+2]}}} , the common ratio of consecutive terms is
{{{r=a[n+2]/a[n+1]=a[n+2]/(-4a[n+2])=-1/4}}} .
It is a geometric sequence, where each term is the one before multiplied by {{{r=-1/4}}} ,
so as for all geometric sequences, {{{a[n]=a[1]*r^(n-1)}}}.
In particular,
{{{a[4]=a[1]*r^(4-1)=a[1]*r^3}}} .
Since {{{a[1]=2}}} and {{{r=-1/4}}} ,
{{{a[4]=2*(-1/4)^3=2*(-1/64)=highlight(-1/32)}}} .
 
The {{{x[n]}}} terms are iterations using {{{f(x)=3x+5}}}
{{{x[0]=1}}}
{{{x[1]=3*x[0]+5=3*1+5=3+5=8}}}
{{{x[2]=3*x[1]+5=3*8+5=24+5=29}}}
{{{x[3]=3*x[2]+5=3*29+5=87+5=highlight(92)}}}
Because the problem asked for the term {{{x[3]}}} , with index {{{3}}},
it was easy enough to calculate each of the terms from the previous term until I got to {{{x[3]}}} .
No need to get more complicated.
If the problem had asked for a term with a much higher index, such as the 10th term,
I would have found a "formula" to get that term directly, without calculating all the terms in between.
{{{x[1]=3*x[0]+5}}}
{{{x[2]=3(3*x[0]+5)+5=x[0]*3^2+5*3+5}}}
{{{x[3]=3(x[0]*3^2+5*3+5)+5=x[0]*3^3+5*3^2+5*3+5}}}
{{{"......................."}}}
{{{x[n]=x[0]*3^n+5*3^(n-1)+5*3^(n-2)+"..."+5*3+5}}}
{{{x[n]=x[0]*3^n+5(3^(n-1)+3^(n-2)+"..."+3+1)}}}
Since {{{3^(n-1)+3^(n-2)+"..."+3+1}}} is the sum of a geometric sequence with first term {{{1}}} and common ratio {{{r=3}}} ,
that sum is {{{3^(n-1)+3^(n-2)+"..."+3+1=(3^n-1)(3-1)}}} , so
{{{x[n]=x[0]*3^n+5((3^n-1)/(3-1))}}}
{{{x[n]=x[0]*3^n+5(3^n-1)/2)}}}
That last "formula" applied to {{{n=3}}} , and with {{{x[0]=1}}} , yields
{{{x[3]=1*3^3+5(3^3-1)/2)=27+5*(27-1)/2=57+5*26/3=27+5*13=27+65=92}}} .