Question 1116736
<pre>
The good lady mistakenly thought it was an arithmetic 
sequence, but it is a quadratic sequence, so we assume:

{{{matrix(1,3,

a[n],""="",An^2+Bn+C)}}}

[Don't confuse the capital A with the subscripted small a<sub>n</sub>,
for the nth term.  They are different.]

There are 3 unknowns A,B, and C, so we need 3 equations,
so we only need the first 3 terms, with n=1,2,3.

{{{a[1]=6}}}, {{{a[2]=12}}}, {{{a[3]=20}}}

Substitute n=1:

{{{matrix(1,3,

a[1],""="",A(1)^2+B(1)+C)}}}

{{{matrix(1,3,

6,""="",A+B+C)}}}

So the first equation is {{{A+B+C=6}}}

Substitute n=2:

{{{matrix(1,3,

a[2],""="",A(2)^2+B(2)+C)}}}

{{{matrix(1,3,

12,""="",4A+2B+C)}}}

So the second equation is {{{4A+2B+C=12}}}

Substitute n=3:

{{{matrix(1,3,

a[3],""="",A(3)^2+B(3)+C)}}}

{{{matrix(1,3,

20,""="",9A+3B+C)}}}

So the third equation is {{{9A+3B+C=20}}}

We have this 3×3 system:

{{{system(A+B+C=6,4A+2B+C=12,9A+3B+C=20)}}}

Subtract the first equation from the second:

{{{3A+B=6}}}

Subtract the second equation from the third equation:

{{{5A+B=8}}}

So we have this 2×2 system:

{{{system(3A+B=6,5A+B=8)}}}

Subtract the first equation from the second:

{{{2A=2}}}

{{{A=1}}}

Substitute in

{{{3A+B=6}}}
{{{3(1)+B=6}}}
{{{3+B=6}}}
{{{B=3}}}

Substitute in 

{{{A+B+C=6}}}
{{{1+3+C=6}}}
{{{4+C=6}}}
{{{C=2}}}

We substitute in the expression for the nth term 

{{{matrix(1,3,

a[n],""="",An^2+Bn+C)}}}


{{{matrix(1,3,

a[n],""="",1n^2+3n+2)}}}

{{{matrix(1,3,

a[n],""="",n^2+3n+2)}}}    

If you like you can factor that as

{{{matrix(1,3,

a[n],""="",(n+1)(n+2))}}}    

Checking to make sure it works for n=1,...,7 with the given sequence:

{{{matrix(1,7,a[1],""="",(1+1)(1+2),""="",(2)(3),""="",6)}}} 
{{{matrix(1,7,a[2],""="",(2+1)(2+2),""="",(3)(4),""="",12)}}} 
{{{matrix(1,7,a[3],""="",(3+1)(3+2),""="",(4)(5),""="",20)}}} 
{{{matrix(1,7,a[4],""="",(4+1)(4+2),""="",(5)(6),""="",30)}}} 
{{{matrix(1,7,a[5],""="",(5+1)(5+2),""="",(6)(7),""="",42)}}} 
{{{matrix(1,7,a[6],""="",(6+1)(6+2),""="",(7)(8),""="",56)}}} 
{{{matrix(1,7,a[7],""="",(7+1)(7+2),""="",(8)(9),""="",72)}}} 

Edwin</pre>