Question 1089382
{{{8}}}, {{{14}}}, {{{22}}}, {{{32}}}, {{{44}}} 

find differences:
{{{8}}}.........{{{14}}}.........{{{22}}}.........{{{32}}}.........{{{44}}} 
.....{{{6}}}.............{{{8}}}.........{{{10}}}.........{{{12}}}
..............{{{2}}}.............{{{2}}}.........{{{2}}}-> second differences are constant, the sequence can be described by a quadratic formula of the form:

{{{u[n]=an^2+bn+c}}}


now find coefficients {{{a}}},{{{ b}}}, and {{{c}}}

To find the values of {{{a}}}, {{{b}}} and {{{c}}} consider the first {{{3}}} terms.

Using {{{u[1]=8}}} and {{{n=1}}} gives

{{{8=a+b+c}}}..........eq.1

Using {{{u[2]=14}}} and {{{n=2}}} gives

{{{14=a*2^2+b*2+c}}}
{{{14=4a+2b+c}}}..........eq.2


Using {{{u[3]=22}}} and {{{n=3}}} gives

{{{22=a*3^2+b*3+c}}}
{{{22=9a+3b+c}}}..........eq.3

solve the system:

{{{8=a+b+c}}}..........eq.1
{{{14=4a+2b+c}}}..........eq.2
{{{22=9a+3b+c}}}..........eq.3
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start with {{{8=a+b+c}}}..........eq.1 and solve for {{{c}}}
{{{c=8-b-a}}}..........eq.1a

same with {{{14=4a+2b+c}}}..........eq.2
 {{{c=14-4a-2b}}} .........eq.2b

from eq.1a and eq.2b we have
{{{8-b-a=14-4a-2b}}}

{{{-b-a+4a+2b=14-8}}}

{{{3a+b=6}}}................solve for {{{b}}}
{{{b=6-3a}}}...........eq.3c

go to {{{c=8-b-a}}}..........eq.1a, substitute {{{b}}}


{{{c=8-(6-3a)-a}}}
{{{c=8-6+3a-a}}}
 {{{c= 2+2a}}}.........eq.4d

take eq3c and eq.4d, and substitute it in 

{{{22=9a+3b+c}}}..........eq.3
{{{22=9a+3(6-3a)+2+2a}}}
{{{22=9a+18-9a+2+2a}}}
{{{22=20+2a}}}
{{{22-20=2a}}}
{{{2=2a}}}
{{{a=1}}}

find  {{{b=6-3a}}}...........eq.3c
{{{b=6-3*1}}}
{{{b=3}}}

and  {{{c= 2+2a}}}.........eq.4d will be
 {{{c= 2+2*1}}}
 {{{c= 4}}}

so, the nth term for the following progression is:

{{{u[n] = n^2 + 3n + 4}}}  (for all terms given)