Bashing unavailable - part 1

$(a+b+c)^2 = 1= a^2+b^2+c^2 +2(ab+bc+ac)=2 +2(ab+bc+ca)$

$ab+bc+ca = \frac{-1}{2}$

$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ ..... $\color{#D61F06}{identity}$

$3-3abc= 2+\frac{1}{2} \implies abc = \frac{1}{6}$

$(ab+bc+ca)^2= a^2b^2+b^2c^2+c^2a^2+2(ab^2c+abc^2+a^2bc)$

$\frac{1}{4} = a^2b^2+b^2c^2+c^2a^2 +2abc(a+b+c)$

$= a^2b^2+b^2c^2+c^2a^2 +\frac{1}{3}$

Hence $a^2b^2+b^2c^2+c^2a^2 = \frac{-1}{12}$

Now as we want $a^4+b^4+c^4$ we'll square the second expression.

$(a^2+b^2+c^2)^2= a^4+b^4+c^4 +2(a^2b^2+b^2c^2+c^2a^2)$

$4=a^4+b^4+c^4 -\frac{1}{6}$ ....as we obtained that value $\frac{-1}{12}$

Hence $a^4+b^4+c^4=4+ \frac{1}{6} = \frac{25}{6}$

Hence answer is $\boxed{31}$

I guess this solution gives the method to the whole series.... Try it out ! Good Luck

Using Newton's Identity where:

$a_nS_1 + a_{n-1} = 0 \\ a_nS_2 + a_{n-1}S_1 + 2a_{n-2} = 0 \\ a_nS_3 + a_{n-1}S_2 + a_{n-2}S_1 + 3a_{n-3} = 0 \\ a_n S_4 + a_{n-1}S_3 + a_{n-2}S_2 + a_{n-3}S_1 + 4a_{n-4} = 0$

Where we assume that $a_n$ is the leading coefficient of polynomial $P(x)=a_3x^3 + a_2x^2 + a_1x + a_0$ .

After algebraic manipulations, we get that $S(4) = 25/6$ by $a(n-4) = 0$ since the polynomial is of third-degree and can hold only 4 coefficients.

As #Aditya Raut said, a^4+b^4+c^4 = 4 +1/6 = 25/6
thus,
a1 + a2 = 25 +6 = \boxed{31}

a^4+b^4+c^4=(a^2+b^2+c^2)^2- 2(a^2b^2+b^2c^2+c^2a^2) =4-2(a^2b^2+b^2c^2+c^2a^2) so i am in need of the value 2(a^2b^2+b^2c^2+c^2a^2). using (a+b+c)^3=1, I get abc =1/6.

and using the formula x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)

i get 2(a^2b^2+b^2c^2+c^2a^2)= -1/6 and a^4+b^4+c^4=4+1/6=25/6....... and the sum is 31..... thanks