Question 984368
I believe what you really need to calculate is 
{{{S(n)=3*1^2+5*2^2+7*3^2+"..."+(2n+1)*n^2}}}
{{{S(n)=3*1^2+5*2^2+7*3^2+"..."+(2n+1)*n^2=highlight((1/2)n^4+(4/3)n^3+n^2+(1/6)n)}}}


Once you figure out that formula,
{{{S(n)=(1/2)n^4+(4/3)n^3+n^2+(1/6)n}}} can be proven by induction, like this:
1) The formula {{{S(n)=(1/2)n^4+(4/3)n^3+n^2+(1/6)n}}} is true for {{{n=1}}} .
For {{{n=1}}} , {{{S(1)=3*1^2=3}}}
and the formula {{{S(n)=(1/2)n^4+(4/3)n^3+n^2+(1/6)n}}} ,
which tells us that
{{{S(1)=1/2+4/3+1+1/6=3/6+8/6+1+1/6=1+(3+8+1)/6=1+12/6=1+2=3}}} ,
which makes the formula true tor {{{n=1}}} .
2) We can prove that if {{{S(n)=(1/2)n^4+(4/3)n^3+n^2+(1/6)n}}} were true for any {{{n=k}}} whole number, it would be true for {{{n=k+1}}} .

Term number {{{k+1}}} is
{{{(2(k+1)+1)*(k+1)^2=(2k+2+1)(k^2+2k+1)
=(2k+3)(k^2+2k+1)=2k^3+3k^2+4k^2+6k+2k+3=2k^3+7k^2+8k+3}}} .
If {{{S(k)=(1/2)k^4+(4/3)k^3+k^2+(1/6)k}}} , then
{{{S(k+1)=S(k)+2k^3+7k^2+8k+3=(1/2)k^4+(4/3)k^3+k^2+(1/6)k+2k^3+7k^2+8k+3}}}
={{{(1/2)k^4+(4/3+2)k^3+4k^2+(1/6+8)k+3
=(1/2)k^4+(10/3)k^3+8k^2+(49/6)k+3}}}
and that show that the formula is true for {{{n=k+1}}} , because the formula says that
{{{S(k+1)=(1/2)(k+1)^4+(4/3)(k+1)^3+(k+1)^2+(1/6)(k+1)}}}
={{{(1/2)(k^4+4k^3+6k^2+4k+1)+(4/3)(k^3+3k^2+3k+1)+k^2+2k+1+(1/6)(k+1)}}}
={{{(1/2)k^4+2k^3+3k^2+2k+1/2 +(4/3)k^3+4k^2+4k+4/3+k^2+2k+1+(1/6)k+1/6}}}
={{{(1/2)k^4+(4/3+2)k^3+8k^2+(8+1/6)k+1/2+4/3+1+1/6}}}
={{{(1/2)k^4+(10/3)k^3+8k^2+(49/6)k+3}}}


HOW WE CAN GET TO THAT FORMULA:
The sum of terms are of the form {{{n^p}}} with {{{p>=1}}} is a polynomial of degree {{{p+1}}} in {{{n}}} with no independent term.
{{{1+2+3+"..."+(n-2)+(n-1)+n=n(n+1)/2=(1/2)n^2+(1/2)n}}} ,
{{{1^2+2^2+3^2+"..."+(n-2)^2+(n-1)^2+n^2=(1/3)n^3+(1/2)n^2+(1/6)n}}} , and so on.
So, the sum of terms of the form {{{2n+1)n^2=2n^3+n^2}}} should be
{{{S(n)=A*n^4+B*n^3+C*n^2+D*n}}}
Knowing that
{{{S(1)=3*1^2=3*1=3}}} ,
{{{S(2)=3*1^2+5*2^2=3+5*4=3+20=23}}} ,
{{{S(3)=3*1^2+5*2^2+7*3^2=23+7*9=23+63=86}}} , and
{{{S(4)=3*1^2+5*2^2+7*3^2+9*4^2=86+9*16=86+144=230}}} ,
we can find {{{A}}} , {{{B}}} , {{{C}}} , and {{{D}}} by solving
{{{system(A+B+C+D=3,16A+8B+4C+2D=23,81A+27B+9C+3D=86,256A+64B+16C+4D=230)}}}