Bashing unavailable - part 6

$We\quad have\\ ab+bc+ac=\frac { 1 }{ 2 } \\ \quad \quad abc=\frac { 1 }{ 6 } \\ Let\quad \\ { A=a }^{ n }+{ b }^{ n }+{ c }^{ n }\\ B={ a }^{ n-1 }+{ b }^{ n-1 }+{ c }^{ n-1\quad }\\ C={ a }^{ n-2 }+{ b }^{ n-2 }+{ c }^{ n-2 }\\ Then\quad \\ A=({ a }^{ n }+{ b }^{ n }+{ c }^{ n })(a+b+c)\\ as\quad a+b+c=1\\ A={ a }^{ n+1 }+a{ b }^{ n }+a{ c }^{ n }\\ \quad +b{ a }^{ n }+{ b }^{ n+1 }+b{ c }^{ n }\\ \quad +c{ a }^{ n }+c{ b }^{ n }+{ c }^{ n+1 }\\ Let\\ D={ a }^{ n+1 }+{ b }^{ n+1 }+{ c }^{ n+1 }\\ Then\\ A=D+ab({ a }^{ n-1 }+{ b }^{ n-1 })+\\ \quad \quad \quad \quad \quad bc({ b }^{ n-1 }+{ c }^{ n-1 })+\\ \quad \quad \quad \quad \quad ac({ a }^{ n-1 }+{ c }^{ n-1 })\\ A=D+ab(B-{ c }^{ n-1 })\\ \quad \quad \quad \quad +bc(B-{ a }^{ n-1 })\\ \qquad \quad +ac(B-{ b }^{ n-1\quad })\\ A=D+B(ab+bc+ac)-abc({ a }^{ n-2 }+{ b }^{ n-2 }+{ c }^{ n-2 })\\ A=D-\frac { B }{ 2 } -\frac { C }{ 6 } \quad \\ D=A+\frac { B }{ 2 } +\frac { C }{ 6 } \\ Hence\quad We\quad have\quad recursive\quad formula\quad for\quad calculating\quad \\ f(n+1)={ a }^{ n+1 }+{ b }^{ n+1 }+{ c }^{ n+1 }\quad in\quad terms\quad of\quad \\ f(n),f(n-1)\quad and\quad f(n-2).\quad$

Using Newton's Identity...

But since the polynomial is of third-degree polynomial, there are 4 terms and since there are 4 terms, there are 4 numerical coefficients, so for a(n-k) = 0 for k greater than or equal to 4.

This reduces to a(n)S(7) + a(n-1)S(6) + a(n-2)S(5) + a(n-3)S(4) = 0 where from earlier, we assumed that a(n) = 1, so a(n-1) = -1, a(n-2) = -1/2, and a(n-3) = -1/6.

Where we assume that a(n) = leading coefficient of polynomial = 1 of a(3)x^3 + a(2)x^2 + a(1)x + a(0) = P(x).

After algebraic manipulations, we get that S(9) = 905/36 and from earlier, S(8) = 1265/72, S(7) = 221/18, S(6) = 103/12, S(5) = 6, and S(4) = 25/6. Hence, the answer.

$Using~Newton~ and~notations,~from~ the~ given~ data:-\\ P_1~=~S_1=1...............................................................................P_1=1\\ P_2~=~S_1*P_1 - 2*S_2~~~=~2,~so......S_2~=~-1/2.......................P_2=2\\ P_3~=~S_1*P_2 - S_2*P_1~+3*S_3*=~3,~so......S_3~=~1/6..................P_3=3\\ ~~~~\\ For~Greater~than~P_3,~~P_i~=~S_1*P_{i-1} - S_2*P_{i-2}+S_3*{i-3}.\\ So ~for~this~problem~~~\color{#3D99F6}{P_i~=~P_{i-1} +1/2*P_{i-2}+1/6*P_{i-3}.}\\ \therefore\\ P_4~=~3+1/2*2+1/6*1~=~25/6=a_1/a_2,.................................a_1=25~~~a_2=6.\\ P_5~=~25/6+1/2*3+1/6*3=6~=~a_3,........................................a_3=6.\\$ $P_6~=6+1/2*25/6+1/6*3=103/12~=~a_4/a_5,.................................a_4=103,~~~a_5=12.\\ P_7~=~103/12+1/2*6+1/6*25/6=~221/18~=~a_6/a_7,.................................a_6=221,~~~a_7=18.\\ P_8~=~221/18+1/2*103/12+1/6*6=~1265/72~=~a_8/a_9,.................................a_8=1265,~~~a_9=72.\\ P_9~=~1265/72+1/2*221/18+1/6*103/12=~905/36~=~a_{10}/a_{11},.................................a_{10}=905,~~~a_{11}=36.\\ \displaystyle \sum_{i=1}^{11}a_i=\Large \color{#D61F06}{2669}.$

use this for this problems. f(n)=f(n-1)+1/2 f(n-2)+1/6 f(n-3) f(n)=a^{n}+b^{n}+c^{n}. Try it