Find the nth term and sum to n terms of the series `12+40+90+168+280+432+"...."`.

Explanation

Let `S_(n)=12+40+90+168+280+432+"....",` then
1st order differences are `28,50,78,112,152,"...."(i.e.Deltat_(1),Deltat_(2),Deltat_(3),".....")`
and 2nd order differences are `22,28,34,40,"...."(i.e.Deltat_(1)^(2),Deltat_(2)^(2),Deltat_(3)^(2),".....")`
and 3rd order dofferences are `6,6,6,6,"...."(i.e.Deltat_(1)^(3),Deltat_(2)^(3),Deltat_(3)^(3),".....")`
and 4th order differences are `0,0,0,0,"...."(i.e.Deltat_(1)^(4),Deltat_(2)^(4),Deltat_(3)^(4),".....")`
`therefore t_(n)=12*^(n-1)C_(0)+28*^(n-1)C_(1)+22*^(n-1)C_(2)+6*^(n-1)C_(3)`
`=12+28(n-1)+(22(n-1)(n-2))/(2)+(6(n-1)(n-2)(n-3))/(1*2*3)`
`=n^(3)+5n^(2)+6n`
and `S_(n)=12*^(n)C_(1)+28*^(n)C_(2)+22*^(n)C_(3)+6*^(n)C_(4)`
`=12n+(28n(n-1))/(2)+(22n(n-1)(n-2))/(1*2*3)+(6*n(n-1)(n-2)(n-3))/(1*2*3*4)`
`=(n)/(12)(n+1)(3n^(2)+23n+46)`.