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When performing binomial expansion of $(a+b\sqrt c)^n$ I get $x+y\sqrt c$ where

  • $x$ is $\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} a^{n-2k} b^{2k} c^k$

  • $y$ is $\sum_{k=0}^{\lfloor (n-1)/2\rfloor} \binom{n}{2k+1} a^{n-2k-1} b^{2k+1} c^k$

Is there a closed end expression for $x$ and $y$? In other words, is there anything that makes summing even/odd binomial coefficients easier?

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Yes: $$x=\frac{\left(a+b \sqrt{c}\right)^n+\left(a-b \sqrt{c}\right)^n}{2}$$ and $$y=\frac{\left(a+b \sqrt{c}\right)^n-\left(a-b \sqrt{c}\right)^n}{2 \sqrt{c}}.$$

In particular, letting $a=b=c=1$, you attain your stated goal of making "summing even/odd binomial coefficients easier": $$\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} =\sum_{k=0}^{\lfloor (n-1)/2\rfloor} \binom{n}{2k+1}=2^n/2.$$

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    $\begingroup$ Not very helpful. That still includes same expression with radicals. $\endgroup$
    – Eugene
    Commented Dec 10, 2021 at 19:03
  • $\begingroup$ What then did you mean, exactly, by "a closed end expression"? What do you want to do with such expressions? The only goal that you clearly stated -- to "make[] summing even/odd binomial coefficients easier" -- is now attained. $\endgroup$
    – Iosif Pinelis
    Commented Dec 10, 2021 at 19:20
  • $\begingroup$ I'd like to be able to quickly calculate expressions like $(3+2\sqrt 5)^{1000}$. Right now I'm using exponentiation by squaring. I was hoping that there's a more efficient way to do the calculation. $\endgroup$
    – Eugene
    Commented Dec 10, 2021 at 19:32
  • $\begingroup$ @Eugene : Well, "to quickly calculate expressions like $(3+2\sqrt 5)^{1000}$" is not at all what is actually asked in your post: "a closed end expression for $x$ and $y$" and "anything that makes summing even/odd binomial coefficients easier". So, who do you think should assume the responsibility for the incorrectly stated question? Also, Mathematica computes $x$ and $y$ for $(3+2\sqrt 5)^{1000}$ in about 0.0003 sec. I am sure there are free-license programs that can do similarly. So, how can that be a problem? $\endgroup$
    – Iosif Pinelis
    Commented Dec 10, 2021 at 20:28
  • $\begingroup$ Sorry if I stated the question in a confusing way. Using straightforward exponentiation by squaring algorithm allows me to calculate that value as fast as Mathematica or even faster. I was wondering if there exists an even faster way to do this using some closed end expression. $\endgroup$
    – Eugene
    Commented Dec 10, 2021 at 23:25
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If the point is a formula that does not include radicals, you may write
$$\begin{bmatrix}x \\[0.3em]y \end{bmatrix}:=\begin{bmatrix}a & bc \\[0.3em]b&a\end{bmatrix}^n\begin{bmatrix}1\\[0.3em]0\end{bmatrix} . $$

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