find $a_1+a_3+a_5+\cdots+a_{37}+a_{39}$

Question

Let $(1+x-2x^2)^{20} = \sum_{r=0}^{40} a_r x^r.\;\;$ Find $$a_1+a_3+a_5+\cdots+a_{37}+a_{39}$$

I mean, at first, I don't have any idea how this can be solved. I tried factorizing the LHS: $((2x-1)(1-x))^{20}$ and then using binomial theorem, I get something like:

$$\Biggl(\binom{20}0(2x)^{20}-\binom{20}1(2x)^{19}+\binom{20}2(2x)^{18}-\binom{20}3(2x)^{17}+\cdots-\binom{20}{19}(2x)+1\Biggr)\Biggl(\binom{20}0(x)^{20}-\binom{20}1(x)^{19}+\binom{20}2(x)^{18}-\binom{20}3(x)^{17}+\cdots-\binom{20}{19}(2x)+1\Biggr)$$

Expanding the RHS:

$$a_0+a_1x^1+a_2x^2+\cdots+a_{39}x^{39}+a_{40}x^{40}$$

But I don't get what to do after this. Taking $x$ common doesn't help, nor in the former equation($2x$).

So how do we do it? A hint would be nice.

Answer 1

If you have a polynomial $f(x)=a_0+a_1x+a_2x^2+\cdots +a_n x^n$, then the sum of the odd indexed coefficients is $$a_1+a_3+\cdots=\frac{f(1)-f(-1)}2.$$ Here you know what $f(x)$ is.

Answer 2

thanks guys, i have solved it too.

$$(1+x-2x^2)^{20} = \sum_{r=0}^{40} a_r x^r$$

putting $x = 1$

$$\sum_{r=0}^{40} a_r=0\;\;\;\;\;\;\;\;\;\;\;\;\;-(1)$$

putting $x = -1$

$$-a_1+a_2-a_3+a_4-a_5+\cdots+a_{40}=2^{20}\;\;\;\;\;\;\;\;\;\;\;\;\;-(2)$$

calculating $(1)-(2)$

$$a_1+a_3+a_5+a_7+\cdots+a_{39}=-2^{19}$$