Can we stop Binomial Summation at any place? (1)

Relevant wiki: Binomial Theorem - Expansions - Basic

Hint

If you guys want, I can upload the full solution, Here is a bit hint,

Use this result

$(1+x)^{-1}=1-x+x^2-x^3+...$

And multiply it with the Binomial that is $(1+x)^{24}$ . Now check for some coefficient which will give you this answer!

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For those, Who tried the hint but couldnt solve .

Here we go,

So here we proceed.

Our Taylor series is $(1+x)^{-1}=1-x+x^2-x^3...$

And $(1+x)^{24}=\dbinom{24}{0} + \dbinom{24}{1} x + .... \dbinom{24}{16} x^{16} + ...+ \dbinom{24}{24}x^{24}$

Now lets find $\displaystyle{\large\sum_{r=0}^{16} (-1)^r \dbinom{24}{r}=?}$

Lets say coeff. of $x^{16} \space \text{in} (1+x)^{24}$ = $\dbinom{24}{16} x^{16}$ .

Now lets multiply $(1+x)^{-1}$ with $(1+x)^{24}$ and try to find out the coeff. of $x^{16}$ , Because while multiplying... $\dbinom{24}{k} x^{16}$ will go with the first term of the taylor series and so on . And if you check the coeff. of $x^{16}$ , You will get the answer.

So Coeff. of $x^{16} \space \text{in} (1+x)^{23}$ = $\dbinom{23}{16}$ .

Now you can find the rest of answer easily!