> First, perhaps I should explain some notation :-) a_11 is the letter `a',
> followed by 11 in subscript. x^2 is the letter `x', followed by the letter
> 2 in superscript (ie. `x^2' would be mathematically the same as `x*x').
> OK, here goes:
>
> If (3x^2 - x - 2)^6 = (a_12)x^12 + (a_11)x^11 + ... + (a_1)x + a_0, what
> is a_0 + a_2 + a_4 + ... + a_12?
Maple, UBasic, ugh. Are we that dependent on computers???
Warning, spoiler follows...
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- define f(x)=(3x^2-x-2)^6=a_0+x*a_1+x^2*a_2+...+x^12*a_12
- Note that f( 1)=0=a_0+a_1+a_2+a_3+...+a_12
- Note that f(-1)=64=a_0-a_1+a_2-a_3+...+a_12
- So, f(1)+f(-1)=64=2*(a_0+a_2+a_4+...+a_12)
- Therefore, the required sum is 32.
-- Tim
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