Question 1069954
DISCLAIMERS:
1) Different teachers/schools/countries use
different symbols/formulas/names. For a few examples, what in USA is usually called arithmetic sequence is an arithmetic progression, or an AP in other places, and "trapezoid" and "billion" also change meaning as you move from one place to another.
2) Your teacher may have given you formulas and "recipes" to solve
"problems of this type" , and may even have a certain preferred or required homework format. I cook from scratch, and solve problems the same way, without using memorized formulas or procedures, only remembering formulas that I have used too often, and seem obvious to me from the most basic definitions.
 
(MY) SYMBOL DEFINITIONS:
{{{x[n]}}}= term number {{{n}}} in the quadratic sequence.
{{{d[n]}}}= first-difference number {{{n}}} in the quadratic sequence, so that
{{{x[n+1]=x[n]+d[n]}}}
FOR THIS SEQUENCE
{{{system(d[1]=1,d[2]=3,d[3]=5)}}} tells us that the second-difference is
{{{2=3-1=5-3}}} throughout the sequence.
The first-differences form an arithmetic sequence
with first term {{{d[1]=1}}] and common difference {{{2}}} , so
{{{d[n]=1+2*(n-1)=1+2n-2=2n-1}}} so {{{highlight(d[n]=2n-1)}}}
 
1. For {{{d[4]}}} ,  {{{d[5]}}} ,  {{{d[6]}}} , you could just count by 2's to get {{{d[4]=7}}} ,' {{{d[5]=9}}} ,' {{{d[6]=11}}} ,
and then find {{{x[5]}}} and {{{x[6]}}} ,
the 5th and 6th terms of the quadratic sequence,
maybe doing mental math while tabulating results, like this
{{{matrix(8,3,n,x[n],d[n],
1,"?",1,2,"?",3,3,"?",5,
4,"?",7,5,"?",9,6,"?",11,
7,35,"-")}}} ---> {{{matrix(8,3,n,x[n],d[n],
1,"?",1,2,"?",3,3,"?",5,
4,"?",7,5,"?",9,6,35-11=highlight(24),11,
7,35,"-")}}} ---> {{{matrix(8,3,n,x[n],d[n],
1,"?",1,2,"?",3,3,"?",5,
4,"?",7,5,24-9=highlight(15),9,6,24,11,
7,35,"-")}}}
or you could apply the formula for {{{d[n]}}} above to get
{{{d[5]=2*5-1=9}}} and {{{d[6]=2*6-1=11}}} .
Either approach allows you to calculate {{{x[5]}}} and {{{x[6]}}} ,
the 5th and 6th terms of the quadratic sequence.
{{{system(x[7]=35,x[7]=x[6]+11)}}}--->{{{system(x[7]=35,35=x[6]+11)}}}--->{{{x[6]=35-11}}}--->{{{highlight(x[6]=24)}}}
{{{system(x[6]=4,x[6]=x[5]+9)}}}--->{{{system(x[6]=24,24=x[5]+9)}}}--->{{{x[5]=24-9}}}--->{{{highlight(x[5]=15)}}}
 
2. Determining the nth term means finding a formula for {{{x[n]}}} ,
and that is where you will show your algebra skills.
You may want to find {{{x[1]}}} .
You could keep working backwards to get {{{x[1]}}} ,
maybe without writing the calculations, just by tabulating the results of your mental math work,
{{{matrix(8,3,n,x[n],d[n],
1,-1,1,2,0,3,3,3,5,
4,8,7,5,15,9,6,24,11,
7,35,"-")}}} ,
or maybe writing down calculations,
{{{x[4]=15-d[4]=15-7=8}}} ,
{{{x[3]=8-d[3]=8-5=3}}} ,
{{{x[2]=3-d[2]=3-3=0}}} ,
{{{x[1]=0-d[1]=0-1=-1}}} ,
or you could calculate {{{x[1]}}} using formulas.
Since the differences form an arithmetic sequence,
we know how to calculate the sum of all the {{{k}}} first-differences
that have to be added to get from {{{x[1]}}} to term number {{{k+1}}}
{{{d[1]+d[2]+"..."+d[k]=k(d[1]+d[k])/2=k(1+2k-1)/2=k(2k)/2=k^2}}}
{{{x[k+1]=x[1]+d[1]+d[2]+"..."+d[k]=x[1]+k^2}}}
so {{{highlight(x[k+1]=x[1]+k^2)}}} is an easy to calculate formula that you can use to relate {{{x[1]}}} to a generic term number {{{k+1}}} .
So {{{x[6+1]=x[1]+6^2}}}
{{{x[7]=x[1]+36}}}
{{{35=x[1]+36}}}
{{{35-36=x[1]}}}
{{{x[1]=-1}}} .
Now, for {{{x[n]}}} , we can use the simple formula {{{x[k+1]=x[1]+k^2}}
figured out above, with {{{n=k+1}}}<--->{{{k=n-1}}}
{{{x[n]=x[1]+(n-1)^2}}}
{{{x[n]=-1+(n^2-2n+1)}}}
{{{highlight(x[n]=n^2-2n)}}}