define S(n)=a^n+b^n+c^n S(1)=1 S(2)=2 S(3)=3 S(4)=4 S(n) satisfies S(n+3)-(a+b+c)S(n+2)+(ab+bc+ca)S(n+1)-abcS(n)=0 Since a+b+c=1, ab+bc+ca=(1/2)((a+b+c)^2-a^2-b^2-c^2)=-1/2, abc=(1/3)(a^3+b^3+c^3-(a+b+c)(a^2+b^2+c^2-ab-bc-ca))=1/6 Therefore S(n+3)=S(n+2)+(1/2)S(n+1)+(1/6)S(n) Put n=2

S(5)=S(4)+(1/2)S(3)+(1/6)S(2)=4+(3/2)+(1/3)=35/6