Solve $U_{n+1}=U_n+2n^2+3n+4, U_1=2$
My problem occurs when trying to find the particular equation, as the variable start to cancel out. This is my working:
Reduced form: $U_{n+1}=U_n$
Auxiliary equation: $r^{n+1}=r^n$, so $r=1$.
$C(n)=A(1^n)=A$
$P(n)=an^2+bn+c$
$a(n+1)^2+b(n+1)+c=an^2+bn+c+2n^2+3n+4$
$an^2+2an+a+bn+b+c=an^2+bn+c+2n^2+3n+4$
Considering coefficients of $n^2$: $a=a+2$ (no solutions)
Considering coefficients of $n$: $2a+b=b+3$, so $a=\frac{3}{2}$.
Considering coefficients of constants: $a+b+c=c+4$, so $b=\frac{5}{2}$
Hence, $P(n)=\frac{3}{2}n^2+\frac{5}{2}n$
General solution: $U_n=A+\frac{3}{2}n^2+\frac{5}{2}n$
Since $U_1=2$, $2=A+\frac{3}{2}+\frac{5}{2}$, so $A=-2$
Final solution: $U_n=\frac{3}{2}n^2+\frac{5}{2}n-2$.
However, this final solution does not correspond to the sequence generated by the initial inductive definition. Could any help me by explaining where my method above went wrong?
$$U_{n+1}-U_n=2n^2+3n+4$$ implies that $U_n=An^3+Bn^2+Cn+D$, then $$U_{n+1}-U_n=3An^2+(3A+2B)n+(A+B+C)=2n^2+3n+4$$ $$\implies A=2/3, 3A+2B=3,A+B+C=4, \implies B=1/2,C=17/6$$ Next given that $U_1=2=A+B+C+D \implies D=-2$. Finally we get $$U_n=\frac{2n^2}{3}+\frac{n^2}{2}+\frac{17n}{6}-2$$