Lucky money new year

$maximum\quad power\quad of\quad n\quad is\quad 2\quad in\quad { u }_{ n }-5{ u }_{ n-1 }+{ u }_{ n-2 }\quad \Rightarrow \left\{ k,l \right\} =\left\{ 2,1 \right\} \\ \\ { u }_{ n }-5{ u }_{ n-1 }+{ u }_{ n-2 }=2{ n }^{ 2 }+2n+1\\ \\ =a{ b }^{ n-2 }\left( { b }^{ 2 }-5b+6 \right) +c{ d }^{ n-2 }\left( { d }^{ 2 }-5d+6 \right) +e\left[ { n }^{ 2 }-5{ \left( n-1 \right) }^{ 2 }+6{ \left( n-2 \right) }^{ 2 } \right] \\ \\ +f\left[ n-5\left( n-1 \right) +6\left( n-2 \right) \right] +\left( g-5g+6g \right) \\ \\ =a{ b }^{ n-2 }\left( { b }^{ 2 }-5b+6 \right) +c{ d }^{ n-2 }\left( { d }^{ 2 }-5d+6 \right) +e\left( 2{ n }^{ 2 }-14n+19 \right) +f\left( 2n-7 \right) +2g\\ \\ compare\quad coefficients\quad of\quad n,\\ \\ \left( { b }^{ 2 }-5b+6 \right) =\left( { d }^{ 2 }-5d+6 \right) =0;\quad \Rightarrow \left\{ b,d \right\} =\left\{ 3,2 \right\} \\ \\ 2e=2,\quad -14e+2f=2,\quad 19e-7f+2g=1\quad \Rightarrow e=1,\quad f=8,\quad g=19\\ \\ put\quad these\quad values\quad in\quad given\quad equations,\\ \\ { u }_{ 0 }=a+c+19=-1;\quad { u }_{ 1 }=3a+2c+28=3\\ \\ \Rightarrow a=15,\quad c=-35\\ \\ \Rightarrow Ans=a+b+c+d+e+f+g+k+l=15+3-35+2+1+8+19+2+1=16$

Relevant wiki: Linear Recurrence Relations - Problem Solving

$u_n = n (n + 8) - 35 .2^n + 5 .3^{n + 1} + 19$

To see how to solve a recurrence relation, see this .For the solution itself, it is just the same as this but see the 1st link or else the second will overwhelm you