Bino-Combi-Turals

k = 0 10 ( 1 ) k ( 20 2 k ) = ? \large \sum_{k=0}^{10} (-1)^{k} \binom{20}{2k} = \, ?

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


The answer is -1024.

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2 solutions

Chew-Seong Cheong
Oct 18, 2016

S = k = 0 20 ( 1 ) k ( 20 2 k ) Since ( 20 2 k ) = 0 for 2 k > 20 = k = 0 10 ( 1 ) k ( 20 2 k ) = ( 20 0 ) ( 20 2 ) + ( 20 4 ) ( 20 6 ) + . . . + ( 20 20 ) \begin{aligned} S & = \sum_{k=0}^{20} (-1)^k {20 \choose 2k} \quad \quad \small \color{#3D99F6}{\text{Since }{20 \choose 2k} = 0 \text{ for }2k > 20} \\ & = \sum_{k=0}^{\color{#3D99F6}{10}} (-1)^k {20 \choose 2k} \\ & = {20 \choose 0} - {20 \choose 2} + {20 \choose 4} - {20 \choose 6} + ... + {20 \choose 20} \end{aligned}

We note that:

( 1 + i ) 20 = ( 20 0 ) + ( 20 1 ) i ( 20 2 ) ( 20 3 ) i + ( 20 4 ) . . . + ( 20 20 ) S = { ( 1 + i ) 20 } = { ( 2 i ) 10 } = { 2 10 ( 1 ) } = 1024 \begin{aligned} (1+i)^{20} & = {20 \choose 0} + {20 \choose 1}i - {20 \choose 2} - {20 \choose 3}i + {20 \choose 4} - ... + {20 \choose 20} \\ \implies S & = \Re \left \{(1+i)^{20} \right \} \\ & = \Re \left \{(2i)^{10} \right \} \\ & = \Re \left \{2^{10}(-1) \right \} \\ & = \boxed{-1024} \end{aligned}

my solution is same too sir !! can we do it by another method ?

A Former Brilliant Member - 4 years, 8 months ago

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I don't know a better method. I thought you used a different one. I also wanted to show a proper LaTex solution. I edited the problem for you.

Chew-Seong Cheong - 4 years, 8 months ago

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thanks sir , You're so nice !

A Former Brilliant Member - 4 years, 8 months ago

plz edit the problem correctly @Chew-Seong Cheong

A Former Brilliant Member - 4 years, 8 months ago

hello shubham dhull

Anubhav Tyagi - 4 years, 7 months ago

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hello @Anubhav Tyagi

A Former Brilliant Member - 4 years, 7 months ago

man missing you on slack

Anubhav Tyagi - 4 years, 7 months ago

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i am too missing slack !

A Former Brilliant Member - 4 years, 7 months ago

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