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How do I solve problems of type :$$\sum_{k=1}^{(n+1)/2}\binom n{2k-1}x^k\text{ or }\sum_{k=0}^{n/2}\binom n{2k}x^k$$

I tried transferring the binomial to $n-1$ but the repeating $x^k$ makes it weird.

Edit: I started with $$(1+x)^n+(1-x)^n$$ with $x=\sqrt5$

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  • $\begingroup$ @robjohn but the power of x will be changed then no? $\endgroup$ – Anvit Apr 13 '18 at 15:19
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    $\begingroup$ I don't see the question, now. Your edit says you started with what seemed to be almost the answer. $\endgroup$ – robjohn Apr 13 '18 at 15:20
  • $\begingroup$ the power of $x$ will be $2k$ in what you started with, so you will need to use $\sqrt{x}$ in the final form. $\endgroup$ – robjohn Apr 13 '18 at 15:21
  • $\begingroup$ I cant really expand $(1+\sqrt5)^{16}+(1-\sqrt5)^{16}$ directly, can I? $\endgroup$ – Anvit Apr 13 '18 at 15:22
  • $\begingroup$ why not? the odd powers of $\sqrt5$ will cancel and you'll be left with twice the even powers of $\sqrt5$, which are integer powers of $5$. $\endgroup$ – robjohn Apr 13 '18 at 15:23
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You can use the binomial theorem twice, $$ (1+x)^n=\sum_{k=0}^n{\binom{n}{k}x^k}\\ (1-x)^n=\sum_{k=0}^n{(-1)^k\binom{n}{k}x^k} $$ If $E$ is the sum of the even terms, and $O$ the sum of the odd terms, we have $$ (1+x)^n=E+O\\ (1-x)^n=E-O $$

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  • $\begingroup$ My bad but that's where i started with $x=\sqrt5$ $\endgroup$ – Anvit Apr 13 '18 at 15:16
  • $\begingroup$ So $E=((1+\sqrt 5)^n+(1-\sqrt 5)^n)/2$ etc. What's the problem? $\endgroup$ – saulspatz Apr 13 '18 at 15:20
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Hint

Consider $$(1+x)^n+(1-x)^n=\sum_{k=0}^n\binom{n}{k}(1+(-1)^k)x^k$$ and $$ (1+x)^n-(1-x)^n=\sum_{k=0}^n\binom{n}{k}(1-(-1)^k)x^k $$ What can you say about $1+(-1)^k$ and $1-(-1)^k$ when $k$ is even or odd?

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  • $\begingroup$ I should've clarified. I started from here and wanted exact value of the term $\endgroup$ – Anvit Apr 13 '18 at 15:14
  • $\begingroup$ @AFalseName: please clarify the question. $\endgroup$ – robjohn Apr 13 '18 at 15:15
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Okay, now that you've added some context, I think I can answer this question. The Binomial Theorem says $$ \left(1+\sqrt{x}\right)^n=\sum_{k=0}^n\binom{n}{k}x^{k/2}\tag1 $$ and $$ \left(1-\sqrt{x}\right)^n=\sum_{k=0}^n\binom{n}{k}(-1)^kx^{k/2}\tag2 $$ Adding $(2)$ to $(1)$ and dividing by $2$ gives $$ \begin{align} \frac12\left(\left(1+\sqrt{x}\right)^n+\left(1-\sqrt{x}\right)^n\right) &=\sum_{k=0}^n\binom{n}{k}\overbrace{\frac{1+(-1)^k}2}^{\text{$1$ when $k$ is even}}x^{k/2}\\ &=\sum_{k=0}^{n/2}\binom{n}{2k}x^k\tag3 \end{align} $$ Subtracting $(2)$ from $(1)$ and dividing by $2$ gives $$ \begin{align} \frac12\left(\left(1+\sqrt{x}\right)^n-\left(1-\sqrt{x}\right)^n\right) &=\sum_{k=0}^n\binom{n}{k}\overbrace{\frac{1-(-1)^k}2}^{\text{$1$ when $k$ is odd}}x^{k/2}\\ &=\sum_{k=1}^{(n+1)/2}\binom{n}{2k-1}x^{k-\frac12}\tag4 \end{align} $$

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