3
$\begingroup$

Use the binomial expansion of$(1+i)^{2n}$ to prove that $$\binom{2n}{0}-\binom{2n}{2}+\binom{2n}{4}-\binom{2n}{6}+...+(-1)^n\binom{2n}{2n}=2^n\cos {\pi n \over 2}, n\in \mathbb{Z^+}$$

I've been trying to solve this problem for quite some time now, but I am unable to make any progress. I first tried to solve it by using De Moivre's theorem which is easy, but not what the question is asking for. Since then I've basically been stuck.

$\endgroup$
3
  • 2
    $\begingroup$ I think the formula should be $$\binom{2n}{0}-\binom{2n}{2}+\binom{2n}{4}-\binom{2n}{6}+...+(-1)^n\binom{2n}{2n}$$$$+i\left(\binom{2n}{1}-\binom{2n}{3}+\binom{2n}{5}-\binom{2n}{7}+...+(-1)^{n-1}\binom{2n}{2n-1}\right)$$$$=2^n(\cos {\pi n \over 2}+i\sin {\pi n \over 2}), n\in \mathbb{Z^+},$$ right? $\endgroup$
    – Paul
    Dec 28, 2011 at 8:26
  • $\begingroup$ @Paul You are right. I guess that I just did not read the question properly. They are looking for the real part of the function and not the imaginary part. Thanks for putting me on the right track! I'll edit the question now. $\endgroup$
    – E.O.
    Dec 28, 2011 at 8:42
  • 2
    $\begingroup$ If the exercise is stated as you wrote, I think you misunderstood the task at hand: what you are asked to do is (i) to use the binomial expansion to recognize that the LHS is the real part of $(1+i)^{2n}$, (ii) to rewrite the complex number $1+i$ in radial form, (iii) to deduce that $(1+i)^{2n}=2^n\mathrm e^{i\pi n/2}$, (iv) to identify the real part of $2^n\mathrm e^{i\pi n/2}$, and (v) to conclude that LHS=RHS. $\endgroup$
    – Did
    Dec 28, 2011 at 9:58

1 Answer 1

Reset to default
3
$\begingroup$

I agree with Paul. The left hand side is an application of your Binomial Theorem while the right hand side comes from De Moivre's Theorem.

In lieu of this, what is true is: $$\binom{2n}{0}-\binom{2n}{2}+\binom{2n}{4}-\binom{2n}{6}+\cdots+(-1)^n\binom{2n}{2n}=2^n\cos {\pi n \over 2}, n\in \mathbb{Z^+}$$ because the real part of the left hand side is equal to the real part of Paul's assertion.

$\endgroup$
1
  • $\begingroup$ You are right. I have changed the question to account for the mistake on my side :D $\endgroup$
    – E.O.
    Dec 28, 2011 at 8:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .