Question 739620
start with
n=1, 2, 3, 4,   5 
to get
1 5 14 28 47 set pairs ({{{x}}},{{{y}}})

{{{x}}}|{{{y}}}

{{{1}}}|{{{1}}}

{{{2}}}|{{{5}}}

{{{3}}}|{{{14}}}

find a line which contains these points


{{{y=ax^2+bx+c}}} plug in first pair

{{{1=a*1^2+b*1+c}}}

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

plug in second  pair

{{{5=a*2^2+b*2+c}}}

{{{5=4a+2b+c}}}......eq.2

plug in third pair

{{{14=a*3^2+b*3+c}}}

{{{14=9a+3b+c}}}......eq.3

you have a system:

{{{1=a+b+c}}}.......eq.1
{{{5=4a+2b+c}}}......eq.2
{{{14=9a+3b+c}}}......eq.3
______________


{{{1=a+b+c}}}.......eq.1...solve for {{{a}}}
{{{1-b-c=a}}}.....plug in eq.2

{{{5=4(1-b-c)+2b+c}}}......eq.2

{{{5=4-4b-4c+2b+c}}}
{{{5=4-2b-3c}}}.......solve for {{{b}}}
{{{2b=4-5-3c}}}
{{{2b=-1-3c}}}
{{{b=-1/2-3c/2}}}

{{{14=9a+3b+c}}}......eq.3...substitute {{{a}}} and {{{b}}}
since {{{a=1-b-c}}}, substitute {{{b}}}
{{{a=1-(-1/2-3c/2)-c}}}

{{{a=1+1/2+3c/2-c}}}
{{{a=3/2+3c/2-2c/2}}}
{{{a=3/2+c/2}}}

{{{14=9(3/2+c/2)+3(-1/2-3c/2)+c}}}.......solve for {{{c}}}

{{{14=27/2+9c/2-3/2-9c/2+c}}}

{{{14=24/2+c}}}

{{{14=12+c}}}

{{{14-12=c}}}

{{{c=2}}}

now find {{{a}}}

{{{a=3/2+c/2}}}

{{{a=3/2+2/2}}}

{{{a=5/2}}}

and find {{{b}}}

{{{b=-1/2-3c/2}}}

{{{b=-1/2-3*2/2}}}

{{{b=-1/2-6/2}}}

{{{b=-7/2}}}

so,

{{{a=5/2}}}
{{{b=-7/2}}}
{{{c=2}}}

since we are looking for general term {{{a[n]}}}, set {{{y=a[n]}}}

{{{a[n]=(5/2)x^2+(7/2)x+2}}}

{{{a[n]=(1/2)(5x^2+7x+2*2)}}}

{{{highlight(a[n]=(1/2) (5n^2-7n+4))}}}....your general term formula where {{{n}}}={{{1}}},{{{2}}},{{{3}}}.......


check: {{{n=1}}}

{{{a[1]=(1/2) (5*1^2-7*1+4)}}}

{{{a[1]=(1/2) (5-7+4)}}}

{{{a[1]=(1/2) (2)}}}

{{{a[1]=1}}}


{{{n=2}}}

{{{a[2]=(1/2) (5*2^2-7*2+4)}}}

{{{a[2]=(1/2) (20-14+4)}}}

{{{a[2]=(1/2) (10)}}}

{{{a[2]=5}}}

and so on.......