1989 AMC 12 Problema 29

Question

Halla

\sum_{k=0}^{49}(-1)^k\binom{99}{2k},

donde

\binom{n}{j}=\frac{n!}{j!(n-j)!}.

Find

\sum_{k=0}^{49}(-1)^k\binom{99}{2k},

where

\binom{n}{j}=\frac{n!}{j!(n-j)!}.

-2^{50}

-2^{49}

0

2^{49}

2^{50}

Reconoce

(-1)^k

como

i^{2k}

Recognize

(-1)^k

as

i^{2k}

Toma la parte real del desarrollo binomial de

(1+i)^{99}

Take the real part of the binomial expansion of

(1+i)^{99}

La suma es la parte real de

(1+i)^{99}=2^{\frac{99}{2}}e^{\frac{99\pi i}{4}}.

Como

99\equiv3\pmod8,

su parte real es

\begin{aligned} 2^{\frac{99}{2}}\cos\frac{3\pi}{4} &=2^{\frac{99}{2}}\left(-\frac{\sqrt2}{2}\right)\\ &=-2^{49}. \end{aligned}

Por tanto, la respuesta correcta es B.

The sum is the real part of

(1+i)^{99}=2^{\frac{99}{2}}e^{\frac{99\pi i}{4}}.

Since

99\equiv3\pmod8,

its real part is

\begin{aligned} 2^{\frac{99}{2}}\cos\frac{3\pi}{4} &=2^{\frac{99}{2}}\left(-\frac{\sqrt2}{2}\right)\\ &=-2^{49}. \end{aligned}

Thus the correct answer is B.

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