Binomial Coeffieients

i = 0 264 ( 1 ) i ( 529 2 i ) \large \displaystyle\sum_{i=0}^{264}(-1)^{i}\binom {529}{2i}

If the value of the summation above is in the form a b a^b , where a a and b b are positive integers with a a is minimized, enter a + b a+b .

Notation : ( M N ) \binom MN denotes the binomial coefficient , ( M N ) = M ! N ! ( M N ) ! \binom MN = \frac{M!}{N!(M-N)!} .


Don't ask why 529.


The answer is 266.

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1 solution

展豪 張
Apr 29, 2016

Relevant wiki: Binomial Theorem

The expression
= ( ( 1 + i ) 529 ) =\Re((1+i)^{529})
= ( ( 2 e i π 4 ) 529 ) =\Re((\sqrt 2 e^{i\frac \pi 4})^{529})
= ( 2 529 / 2 e i π 4 ) =\Re(2^{529/2}e^{i\frac \pi 4})
= 2 529 / 2 ( e i π 4 ) =2^{529/2}\Re(e^{i\frac \pi 4})
= 2 529 / 2 ( cos π 4 + i sin π 4 ) =2^{529/2}\Re(\cos\dfrac\pi 4+i\sin\dfrac\pi 4)
= 2 529 / 2 1 2 =2^{529/2}\dfrac 1{\sqrt 2}
= 2 528 / 2 =2^{528/2}
= 2 264 =2^{264}
The answer = 2 + 264 = 266 =2+264=266


I solved it same way.

Hana Wehbi - 4 years, 9 months ago

Why 529? 😂😂.

Anurag Pandey - 4 years, 8 months ago

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No...don't ask why......

展豪 張 - 4 years, 8 months ago

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