How to calculate the sum $S=\mathop{\sum}\limits_{i=0}^{\lfloor \frac{n}{4}\rfloor}\binom{n}{4i}$ ? Or how to get a good upper bound on $S$ in terms of $n$?
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You might guess that this will be about a quarter of the full binomial sum so $2^{n-2}$ – Henry Oct 08 '19 at 00:51
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@Henry: You might guess, but will it be true?... – darij grinberg Oct 08 '19 at 00:53
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5The closed fomula is $\frac14\left(2^n+(1+i)^n+(1-i)^n+0^n\right)$ here we take $0^0=1.$ – Thomas Andrews Oct 08 '19 at 00:58
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1Calculate it for the first few values of $n$, then look it up in the Online Encyclopedia of Integer Sequences. – Gerry Myerson Oct 08 '19 at 01:02
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1This can be rewritten as $$2^{n-2}+2^{(n-2)/2}\cos(n\pi/4)$$ for $n>0.$ – Thomas Andrews Oct 08 '19 at 01:03
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As we are interested in every fourth term
let us consider $x^4=1,x=\pm i,\pm1$
Let $$a(1+1)^n+b(1-1)^n+c(1-i)^n+d(1+i)^n=S=?$$
Compare the coefficients of $\binom nr$ where $r=4i,4i+1,4i+2,4i+3$ to find
$$a+b+c+d=1$$
$$a-b-ci+di=0$$
$$a+b-c-d=0$$
$$a-b+ci-di=0$$
Add second and fourth $$a-b=0$$
Then subtract second from the fourth
$$i(c-d)$$
From the first $$2b+2d=1$$
From the third $$2b-2d=0$$
$$\implies b=d=\dfrac14$$
lab bhattacharjee
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