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How to do the problem of 20 1 1 Pudong New Area Mathematics Module 2 10?
Let x= 1, then a (0)+a (1)+a (2)+...+a (2n-1)+a (2n) = (1+2) (2n) = 3.

Let x=- 1, then a (0)-a (1)+a (2)-a (3)+...+a (2n-2)-a (2n-1)+a (2n) = (-)

a( 1)+a(3)+...+a(2n- 1)=( 1/2){[a(0)+a( 1)+...+a(2n)]-[a(0)-a( 1)+a(2)-a(3)+...+a(2n)]}

=( 1/2)[3^(2n)- 1]