Let the number of odd coefficients of powers of $x$ in the expansion of

$\displaystyle \sum^{500}_{n=1}(x+1)^n$

be $M$ . Find $2M$

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The answer is 252.

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Note first that $\displaystyle\sum_{n=0}^{500} (x + 1)^{500} = \dfrac{(x + 1)^{501} - 1}{(x + 1) - 1}$ ,

and so $\displaystyle\sum_{n=1}^{500} (x + 1)^{500} = \dfrac{(x + 1)^{501} - 1}{x} - 1 = \dfrac{(x + 1)^{501} - x - 1}{x}$ .

The numerator will then be the same as the expansion of $(x + 1)^{501}$ but with a constant coefficient of $0$ and with the coefficient of $x$ being $500$ , (i.e., even), instead of $501$ .

Thus when this is divided by $x$ , the number of odd coefficients of powers of $x$ will be the number of odd coefficients in the expansion of $(1 + x)^{501}$ minus $2$ , i.e., minus the odd constant and the odd coefficient of $x$ .

Now the coefficients in the expansion of $(x + 1)^{501}$ are of the form $\binom{501}{k}$ with $k$ running from $0$ to $501$ . So the number of odd coefficients in this expansion will be

$\displaystyle\sum_{k=0}^{501} (\dbinom{501}{k}$ (mod $2)) = 128$ ,

and thus $M = 128 - 2 = 126$ , giving us the answer $2M = \boxed{252}$ .

Comment: The rationale behind the mod calculation is that each odd coefficient returns a value of $1$ when evaluated mod $2$ . The actual value was found here using WolframAlpha. I realize that, to some, this may not be a satisfactory approach, but I could not find a pattern when looking at the sequence of binomial coefficients and thus resorted to a bit of WA trickery. If anyone has a better approach it would be most welcome. :)