If (1+x+x^2)^n=a0+a1 x+a2 x^2+.....+a2n x^2n ,then a0+a3+a6+........+=?

Answers (1)

Solution:   Putting   x=1,omega,omega^2  respectively ,

                 3^n=a_0+a_1+a_2+.......+a_2n                                       .........(1)

 

        Rightarrow         (1+omega+omega^2)^n=a_0+a_1omega+a_2omega^2+.......+a_2nomega^2n        .........(2)

       Rightarrow            (1+omega^2+omega^4)^n=a_0+a_1omega^2+a_2omega^4+.......+a_2nomega^4n       .....(3)

      From    (1)+(2)+(3)   we have

       Rightarrow           3^n+(1+omega+omega^2)^n+(1+omega^2+omega^4)^n=3a_0+3a_3+3a_6+...........

     	herefore             3^n-1=a_0+a_3+a_6+....... 

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