Question 924484
{{{4}}},{{{4}}},{{{0}}},{{{-8}}},{{{-20}}}

all given terms are divisible by {{{2}}}, and as you see all are {{{2}}} times the terms {{{2}}},{{{2}}},{{{0}}},{{{-4}}},{{{-10}}}

then we can multiply {{{(3n-n^2)}}} by {{{2}}} and we get

{{{2(3n-n^2)}}}-this is nth term

check if this formula for nth term gives you {{{4}}},{{{4}}},{{{0}}},{{{-8}}},{{{-20}}}

{{{2(3*1-1^2)=2(3-1)=2*2=4}}}

{{{2(3*2-2^2)=2(6-4)=2*2=4}}}

{{{2(3*3-3^2)=2(9-9)=2*0=0}}}

{{{2(3*4-4^2)=2(12-16)=2*(-4)=-8}}}

{{{2(3*5-5^2)=2(15-25)=2*(-10)=-20}}}

but, I like higher degree first, and if you multiply {{{2(3n-n^2)}}} by {{{-1}}}, you get

{{{-2(n^2-3n)}}}-this is nth term too

check:

{{{-2(1^2-3*1)=-2(1-3)=-2(-2)=4}}}

{{{-2(2^2-3*2)=-2(4-6)=-2(-2)=4}}}

{{{-2(3^2-3*3)=-2(9-9)=-2(0)=0}}}

{{{-2(4^2-3*4)=-2(16-12)=-2(4)=-8}}}

{{{-2(5^2-3*5)=-2(25-15)=-2(10)=-20}}}