giúp em dới mọi người ơi

Ta có: `u_n=2u_(n-1)+n^2-n-3 \ forall \ n in NN; n >= 2`

`<=>` `u_n+n^2+3n=2u_(n-1)+2n^2+2n-3`

`<=>` `u_n+n^2+3n=2u_(n-1)+2(n-1)^2+6(n-1)+1`

`<=>` `u_n+n^2+3n=2[u_(n-1)+(n-1)^2+3(n-1)]+1`

Đặt `v_n=u_n+n^2+3n`

`=>` `{(v_1=7),(v_n=2v_(n-1)+1 \ forall \ n in NN; n >= 2):}`

Ta có: `v_n=2v_(n-1)+1 \ forall \ n in NN; n >= 2`

`<=>` `v_n+1=2v_(n-1)+2`

`<=>` `v_n+1=2(v_(n-1)+1)`

Đặt `w_n=v_n+1`

`=>` `{(w_1=8),(w_n=2w_(n-1) \ forall \ n in NN; n >= 2):}`

`=>` `w_n=8*2^(n-1) \ forall \ n in NN^**`

`=>` `w_n=2^(n+2) \ forall \ n in NN^**`

`=>` `v_n=2^(n+2)-1 \ forall \ n in NN^**`

`=>` `u_n+n^2+3n=2^(n+2)-1 \ forall \ n in NN^**`

`=>` `u_n=2^(n+2)-1-n^2-3n \ forall \ n in NN^**`

Ta có: `S_20=u_1+u_2+...+u_20`

`=2^3-1-1^2-3+2^4-1-2^2-3*2+...+2^22-1-20^2-3*20`

`=(2^3+...+2^22)-20-(1^2+2^2+...+20^2)-3*(1+2+...+20)`

`=2^23-8-20-(20*21*41)/6-3*(20*21)/2`

`=8385080`

`=>` Chọn `B`