Question

What is the sum of the coefficients of
( [3x - 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 ) ??

Answer 1

tough one. i'll give it a go.

Answer 2

\[\left(1+3 x-3 x^2\right)^{1489} \]

Answer 3

^ nope... the second term is different to the first term, so you don't just add the indices

Answer 4

=(-(3x^2-3x-1))^744x(3x^2-3x+1)^745
dont forget about
(x+y)(x-y)=(x^2-y^2)
=(3x^2-3x+1)((3x^2-3x)^2-1)^744

Answer 5

Let a = 3x - 3x^2

= (a + 1)^744 * (-a + 1)^745

= (1 + a)^744 * (1 - a)^644 * (1 - a)

= (1 - a^2)^744 * (1 - a)

= (a^2 - 1)^744 * (1 - a)

Now let (a^2 - 1)^744 = M:

and then i'll spare you the details, but i get the coefficients of M can be described
as:

sum(0,1488) 1488Ck * (-1)^k * sum(0,744) 3^(1488-2k) * 744Ck * (-1)^k

is there a simpler way ? This is so messy.

Answer 6

^ nope... the second term is different to the first term, so you don't just add the indices

Answer 7

Yes, an error.

Answer 8

yucky problem lol

Answer 9

The answer is 1.\[\left(1+3 x-3 x^2\right) \left(1-3 x+3 x^2\right)^2= \]\[1-3 x-6 x^2+45 x^3-90 x^4+81 x^5-27 x^6 \]If x is set to 1, the sum of the coefficients is 1.\[\left(1+3 x-3 x^2\right)^{744} \left(1-3 x+3 x^2\right)^{745}\text{/. }x\to 1\text{ }\text{ is}\text{ }1 \]Mathematica Home Edition was used for the calculations.

Answer 10

i think almost all of the middle terms will actually cancel out...let me double check..

Answer 11

An expanded Mathematica solution with comments is attached.