STEP I 1990 question 2 solution

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STEP I 1990 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1990 question 2 solution

If $\omega$ = $e^{\frac{2i\pi}{3}$ , then $\omega$ is a complex cube root of 1 (by DeMoivre's), so $\omega^{3} - 1 = 0 \Rightarrow (\omega - 1)(\omega^{2} + \omega + 1) = 0 \Rightarrow \omega^{2} + \omega + 1 = 0$ as $\omega$ is complex.

$1 + \omega^{2} = -\omega$

$|-\omega| = 1$

$\arg(-\omega) = \frac{-\pi}{3}$

$(1 + \omega)^{n} = (-\omega^{2})^{n}$

$\binom{n}{0} + \binom{n}{1}\omega + \cdots + \binom{n}{n} = (-1)^{n}\omega^{2n}$

$(1 + \omega^{2})^{n} = (-\omega)^{n}$

$\binom{n}{0} + \binom{n}{1}\omega^{2} + \cdots + \binom{n}{n}\omega^{2n} = (-1)^{n}\omega^{n}$

$(1 + \omega^{3})^{n} = 2^{n}$

$\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^{n}$

Note that if you add the three together, the coefficients of all the binomial coefficients, $\binom{n}{k}$ , except for those where k is a multiple of three, are equal to $\omega^{2} + \omega + 1 = 0$ . When k is a multiple of three, the coefficient is three.

So $\binom{n}{0} + \binom{n}{3} + \cdots + \binom{n}{k} = \frac{1}{3}(2^{n} + 2\cos\frac{n\pi}{3})$

As $(-1)^{n}e^{\frac{i2n\pi}{3}} + (-1)^{n}e^{\frac{i4n\pi}{3}} = 2\cos\frac{n\pi}{3}$ by the sum to product formulae.

$\binom{25}{0} + \binom{25}{3} + \cdots + \binom{25}{24} = \frac{1}{3}(2^{25} + 2\cos\frac{\pi}{3}) = 11,184,811$ .

$(\cos\frac{25\pi}{3} = \cos\frac{\pi}{3})$

$\binom{24}{2} + \binom{24}{5} + \cdots + \binom{24}{23} = \frac{1}{2}\left(2^{24} - \frac{1}{3}(2^{24} + 2)\right) = \frac{1}{3}(2^{24} - 1) = 5,592,405$ .