Question

Binomial expansion.

Answer 1

The sum of coefficients of integral powers of \(x\) in the binomial expansion of


\(\large \color{black }{\begin{align}
&&&&&&&&&&&&&&&&&\left(1-2\sqrt x\right)^{50}\hspace{.33em}\\~\\
\end{align}}\)

is

Answer 2

\(\large \color{black }{\begin{align}
&a.)\quad \dfrac{3^{50}-1}{2}\hspace{.33em}\\~\\
&b.)\quad \dfrac{2^{50}+1}{2}\hspace{.33em}\\~\\
&c.)\quad \dfrac{3^{50}+1}{2}\hspace{.33em}\\~\\
&d.)\quad \dfrac{3^{50}}{2}\hspace{.33em}\\~\\
\end{align}}\)

Answer 3

I worked through a lot of complicated stuff and didn't really get to an answer but I think just by looking at the answer choices it should be b simply because I can't see why it would be a power of 3 at all haha. I'm still working on finding a better answer though I can show what I've done so far.

Answer 4

I could be completely wrong with my last post, so here's some actual content. So by applying the binomial theorem\[\Large (1-2 \sqrt{x})^{50}=\sum_{n=0}^{50} \binom{50}{n}(-2 \sqrt{x})^n\] then we can make a substitution n=2m so that we will have integral powers of x terms only. \[\Large \sum_{m=0}^{25} \binom{50}{2m}4^m x^m\] Now you can either plug in x=1 or just get rid of the x terms... Whichever makes you more comfy, they are effectively the same thing, and now we have at least a summation expressing exactly what you asked for: \[\Large \sum_{m=0}^{25} \binom{50}{2m}4^m \] A possibly simplification would be to tear down the binomial, just an idea, but I'm sorta stuck on it at the moment of how to evaluate this.\[\Large 50! \sum_{m=0}^{25} \frac{4^m }{(2m)!(50-2m)!}\] Or another thought I was having was to try to make it more symmetrical on the powers of 4 to match the denominators with negative exponents. But that might be overcomplicating it somehow. Anyways hopefully this helps a bit.

Answer 5

\[\Large \sum_{m=0}^{25} \binom{50}{2m}4^m = \frac{1}{2}(4-1)^{50}\]

Answer 6

+1/2 i think
so im betting on option c

Answer 7

i should inform that i don't know the correct answer

Answer 8

http://www.wolframalpha.com/input/?i=%28%284-1%29%5E50%29%2F2+-+%5Csum%5Climits_%7Bk%3D1%7D%5E%7B25%7D+%28%2850+choose+%282*k%29%292%5E%282*k%29%29+

Answer 9

The answer given is \(C.\) :D