Anyone know how to go about solving this? I've tried adding the individual terms with no luck, and also I don't get how the 'n' in 15Cn doesn't correspond to (-1)^n like in most binomial formats.
$$\sum_{n=0}^5(-1)^n\binom5nx^{5-2n}=32$$
Terribly sorry for the formatting, I'm new here and still trying to figure everything out. Thanks!
On
$$32=\binom50x^{5}-\binom52x^{3} + \binom52x^{1}-\binom53x^{-1} + \binom54x^{-3}-\binom55x^{-5} $$
Or $$32x^5=\binom50x^{10}-\binom52x^{8} + \binom52x^{6}-\binom53x^{4} + \binom54x^{2}-\binom55 $$
$$ 32x^5 = (x^2-1)^5\implies 2x= x^2-1 \implies x=...$$
On
Factoring out the $x^5$ (because it doesn't depend on $n$) we obtain $$x^5\sum_{n=0}^5(-1)^n{5 \choose n}x^{-2n}=32.$$ Now remark that $x^{-2n}=(\tfrac{1}{x^2})^n$. So by the binomial theorem we have $$x^5(1-\tfrac{1}{x^2})^5=32.$$ Recalling that $2^5=32$ we can take the 5th root of both sides to obtain $x(1-\tfrac{1}{x^2})=2$. Finally we multiply both sides by $x$ and rearrange to obtain the equation $$x^2-2x-1=0.$$ I'll leave the rest to you.
It can be written as $$x^{-5}\sum_{n=0}^5 \binom{5}{n}(-1)^n x^{10-2n}$$ $$x^{-5}\sum_{n=0}^5 \binom{5}{n}(-1)^n (x^2)^{5-n}$$ $$=x^{-5}(x^2-1)^5$$ $$=(\frac{x^2-1}{x})^5=32$$
which then can be easily solved.