Question 1187173


 {{{0}}},{{{6}}},{{{14}}},{{{24}}},{{{36}}},{{{50}}},.......


find differences
{{{0}}} ...  {{{6}}}  ... {{{14}}}  ... {{{24}}}  ... {{{36}}}  ... {{{50}}}
 ... {{{6}}} ...  {{{8 }}}... {{{ 10}}} ...  {{{12}}} ...  {{{14}}}  
 ......{{{2}}} ......{{{ 2}}} ......{{{2}}} ......{{{2}}}  


second differences are same, nth term formula is quadratic


{{{a[n]=a*n^2 + b*n +c}}}


{{{a[1]=0}}}
{{{a[2]=6}}}
{{{a[3] = 14}}}


{{{0=a*1^2 + b *1 +c}}}
{{{0=a+ b  +c}}}..........solve for {{{c}}}
{{{c=-a-b}}}....eq.1


{{{6=a*2^2 + b *2 +c}}}
{{{6=4a+ 2b +c}}}............solve for {{{c}}}
{{{c= 6-4a-2b}}}...........eq.2

{{{14=a*3^2 + b *3 +c}}}
{{{14=9a+ 3b +c}}}............solve for {{{c}}}
{{{c= 14-9a-3b}}}...........eq.3


from eq.1 and eq.2 we have
{{{-a-b=6-4a-2b}}}.......solve for {{{b}}}
{{{2b-b=6-4a+a}}}
{{{b=6-3a}}} ..............eq.1))


from eq.1 and eq.3 we have

{{{-a-b=14-9a-3b}}}.....solve for {{{b}}}
{{{3b-b=14-9a+a}}}
{{{2b=14-8a}}}.........divide by {{{2}}}
{{{b=7-4a}}}............eq.2))


equal eq.1)) and eq.2))

{{{6-3a=7-4a}}}....solve for{{{ a}}}
{{{4a-3a=7-6}}}
{{{a=1}}}


go to

{{{b=6-3a}}} ..............eq.1), substitute {{{a}}}
{{{b=6-3*1 }}}
{{{b=3}}}

go to

{{{c=-a-b}}}....eq.1, substitute {{{a }}}and {{{b}}}
{{{c=-1-3}}}
{{{c=-4}}}

{{{n}}}th term formula is:

{{{a[n] =n^2 + 3 n - 4}}}

or

{{{a[n] = (n - 1) (n + 4) }}}