Bashing unavailable - part 6

Algebra Level 4

Given that

a+b+c=1\color{#20A900}{a+b+c =1}

a2+b2+c2=2\color{#3D99F6}{a^2+b^2+c^2=2}

a3+b3+c3=3\color{#D61F06}{a^3+b^3+c^3=3}

Then a4+b4+c4=a1a2a^4+b^4+c^4= \frac{a_1}{a_2} .... where a1a_1 and a2a_2 are positive coprime integers.

And a5+b5+c5=a3a^5+b^5+c^5=a_3 ...... where a3a_3 is an integer.

And a6+b6+c6=a4a5a^6+b^6+c^6= \frac{a_4}{a_5} ....where a4a_4 and a5a_5 are positive coprime integers.

And a7+b7+c7=a6a7a^7 +b^7+c^7= \frac{a_6}{a_7} ....where a6a_6 and a7a_7 are positive coprime integers.

And a8+b8+c8=a8a9a^8+b^8+c^8 = \frac{a_8}{a_9} ... where a8a_8 and a9a_9 are positive coprime integers.

And a9+b9+c9=a10a11a^9+b^9+c^9 = \frac{a_{10}}{a_{11}} ... where a10a_{10} and a11a_{11} are positive coprime integers.

Find a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+a11a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8 +a_9+a_{10}+a_{11}.

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