2
$\begingroup$

Find the sum of the series $$\binom{4n}{0}+\binom{4n}{4}+\binom{4n}{8}+\ldots+\binom{4n}{n}=\sum_{j=0}^{n}\binom{4n}{4j}.$$

My approach is to consider $(1+x)^{4n} = \sum_{j=0}^{4n}\binom{4n}{j}x^j.$

How to proceed further now? Please help on this since I am quite clueless. Thanks a lot .

$\endgroup$
3
$\begingroup$

HINT:

Set $x^4=1$ i.e., $x=\pm1,\pm i$ in the given identity and add

$\endgroup$
3
$\begingroup$

A standard trick to evaluate such binomial sums is to use the discrete Fourier transform (DFT).

The characteristic function $\chi$ of the integers that are multiples of four can be written as: $$ \chi(n) = \frac{1}{4}\left(1^n + i^n + (-1)^n + (-i)^n\right),$$ hence: $$\sum_{j=0}^{n}\binom{4n}{4j} = \sum_{j=0}^{4n}\binom{4n}{j}\chi(j) = \frac{1}{4}\left(2^{4n}+(1+i)^{4n}+(1-i)^{4n}\right) = \color{red}{\frac{16^n+2(-4)^n}{4}}.$$

$\endgroup$
1
$\begingroup$

$$\newcommand{\c}{\binom} (1+x)^{4n}=\c{4n}0+\c{4n}1x+\c{4n}2x^2+....\c{4n}{4n}x^{4n} $$ Now one property of $j$-th roots of unit is(see here for proof): $$\sum_{\alpha}\alpha^k=\begin{cases}0&k\not\equiv0\pmod j\\1&k\equiv0\pmod j\end{cases}$$ So for fourth roots of unit, viz. $\alpha,\beta,\gamma,\delta\leftrightarrow 1,-1,+i,-i$ $$(1+\alpha)^{4n}+(1+\beta)^{4n}+(1+\gamma)^{4n}+(1+\delta)^{4n}=\sum_{k=0}^{4n}\c{4n}{4k}$$

$\endgroup$
  • 2
    $\begingroup$ Does $1$ satisfy $1+x+x^2+x^3=0?$ $\endgroup$ – lab bhattacharjee Jan 11 '15 at 16:12
  • $\begingroup$ @labbhattacharjee: do you mean $x=1$? $\endgroup$ – Marc van Leeuwen Jan 11 '15 at 16:30
  • $\begingroup$ @MarcvanLeeuwen, Yes, definitely. $\endgroup$ – lab bhattacharjee Jan 11 '15 at 16:31
  • $\begingroup$ @labbhattacharjee edited $\endgroup$ – RE60K Jan 11 '15 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.