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Question: What is the integral part of the following expression?$$(a+\sqrt{b})^{2n}+(a-\sqrt{b})^{2n}$$

The question has specific values of $a=2,b=5$ and $2n=2016$.

I was able to simplify (or complexify) it to:$$\sum_{i=0}^n\binom{2n}{2i}a^{2i}b^{n-i}.$$

I think I need to use $$\binom{2n}{2i}=\binom{2n-1}{2i-1}+\binom{2n-1}{2i},$$

but because the powers of $a$ and $\sqrt{b}$ dont change I cant make a closed form.

Please help.

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  • $\begingroup$ yes, it is. but not sure how to calculate it $\endgroup$
    – Anvit
    Apr 2, 2018 at 8:33
  • $\begingroup$ So the answer should an explicit number? $\endgroup$
    – Ѕааԁ
    Apr 2, 2018 at 8:34
  • $\begingroup$ I think so, not sure. $\endgroup$
    – Anvit
    Apr 2, 2018 at 8:35
  • $\begingroup$ Exact value would be big, very big. $\endgroup$
    – King Tut
    Apr 2, 2018 at 8:37
  • $\begingroup$ Note that $\sqrt{b}^{2n-2i} =\sqrt{b}^{2(n-i)} =(\sqrt{b}^2)^{n-i} =b^{n-i}$, so your sum is already an integer. $\endgroup$
    – Rócherz
    Apr 2, 2018 at 8:38

3 Answers 3

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Note that $\displaystyle 2-\sqrt{5}=\frac{-1}{2+\sqrt{5}}$.

So, $\displaystyle (2-\sqrt{5})^{2n}=\frac{1}{(2+\sqrt{5})^{2n}}\in(0,1)$.

Let $K=(2+\sqrt{5})^{2n}+(2-\sqrt{5})^{2n}$ which is an integer.

$$ (2+\sqrt{5})^{2n}+(2-\sqrt{5})^{2n}= (2+\sqrt{5})^{2n}+\frac{1}{(2+\sqrt{5})^{2n}}$$

$$ K=\lfloor(2+\sqrt{5})^{2n}\rfloor+1$$

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  • $\begingroup$ Well this is an observation, not sure exactly what op wants. $\endgroup$
    – King Tut
    Apr 2, 2018 at 8:42
  • $\begingroup$ But there no obvious way here to calculate K $\endgroup$
    – Anvit
    Apr 2, 2018 at 8:43
  • $\begingroup$ Also, +1 may or may not occur depending on value $\endgroup$
    – Anvit
    Apr 2, 2018 at 8:44
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    $\begingroup$ @AFalseName No Its obviously greater that floor of that so +1 is there. Both are positive terms. $\endgroup$
    – King Tut
    Apr 2, 2018 at 8:45
  • $\begingroup$ I think $a=2$ and $b=5$ is not arbitrary in the question of the admission test. $\frac{1}{(2+\sqrt{4})^{2n}}<\frac{1}{4^{2n}}$ is very small. $K$ is $ (2+\sqrt{5})^{2n}+$ a very small quantity. Since $K$ is an integer, it is $\lfloor (2+\sqrt{5})^{2n}\rfloor+1$ $\endgroup$
    – CY Aries
    Apr 2, 2018 at 8:49
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It is the Lucas number $L_{6048}$.

Explicitly: $$\eqalign{&9062504429942830612498674723940636856073005180754102851103202470654400480862004\cr&470945569990318549786600962729358391458409742980121786528345516555311724482526\cr&469267214426875327939356480512886766136191123865898185909060509390911530211575\cr&250068296151787836515687295491384367878337029779507202772473247457611189758829\cr&765936634113604899116555301784760851901747848131875756210602568727367710580171\cr&895805654908212158642925592215357660076677734346857341254456623431865984405026\cr&892929174068925635521071005404413310101081812578653451740631688019712993398028\cr&457721524261115727978848468969621355164549034177997064126859568812631509397511\cr&016984199127347171026602653375690100545552519714738487571315942718492813756706\cr&082204335257998697204024192225222110769079285160314467805807823101456293333013\cr&846744313345219198803135032572074342248040447170428652954313635095193406457345\cr&312384800612815239265977428989281668436611671607920070840254359108721833323841\cr&188078628899091481176464496313126989415994801739336416785720301217234956031027\cr&978552035253675749937689863092976010498341281125560768668240052298434396165671\cr&537289666399741274176671027424411077085703975089732293442712775771653296648523\cr&231441100434648095729915791965725056427249879681535562607728243485574370058117\cr&109750914437122}$$

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  • $\begingroup$ Indeed, but is this expected in a paper! $\endgroup$
    – King Tut
    Apr 2, 2018 at 9:04
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Define $c_0 = 2$, $c_1 = 4$, and $c_{n + 2} = 4c_{n + 1} + c_n$ for $n \geqslant 0$, then$$ c_n = (2 + \sqrt{5})^n + (2 - \sqrt{5})^n. \quad \forall n \in \mathbb{N} $$

Note that$$ \begin{pmatrix}c_{n + 2}\\c_{n + 1}\end{pmatrix} = \begin{pmatrix}4 & 1\\ 1 & 0\end{pmatrix} \begin{pmatrix}c_{n + 1}\\c_n\end{pmatrix}. $$ Denote $\displaystyle A = \begin{pmatrix}4 & 1 \\ 1 & 0\end{pmatrix}$, then$$ \begin{pmatrix}c_n\\c_{n - 1}\end{pmatrix} = A^{n - 1} \begin{pmatrix}c_1\\c_0\end{pmatrix}. $$ (Tedious computation starts.) Because$$ 2016 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5, $$ and$$ A^2 = \begin{pmatrix}17 & 4 \\ 4 & 1\end{pmatrix}, A^{2^2} = \begin{pmatrix}305 & 72 \\ 72 & 17\end{pmatrix}, A^{2^3} = \begin{pmatrix}98209 & 23184 \\ 23184 & 5473\end{pmatrix}, \cdots $$ then the explicit answer can be computed, but it is too hard to continue computing manually.

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