Let n be an odd integer greater than $$1$$. Prove that the sequence
$$ \binom{n}{1}, \binom{n}{2},...., \binom{n}{\frac{n-1}{2}}$$
contains an odd number of odd numbers.
Even + Odd = Odd
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Re: Even + Odd = Odd
I think the problem is incomplete . Because if $n$ is an an even integer , then the sequence $ \binom{n}{1} , \binom{n}{2} , ....., \binom{n}{\frac n2}$ contains either no odd integer or an odd number of odd integer(s) . Almost similar , isn't it ?
Re: Even + Odd = Odd
@Sakib
But the problem says that $n$ can only be an odd integer greater than $1$. Why are you bothered about even values of $n$?
But the problem says that $n$ can only be an odd integer greater than $1$. Why are you bothered about even values of $n$?
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Re: Even + Odd = Odd
We know, \( { n \choose k} = {n \choose n-k}\).
So, \({ n \choose 1} + { n \choose 2}+ ...... +{ n \choose \frac {n-1}{2}} = \frac {1}{2} \times [{ n \choose 1} + { n \choose 2}+ ...... + { n \choose n-1}] = \frac {1}{2} \times ( 2^n - 2 ) = 2^{n-1} - 1\)
Where, \(2^{n-1} -1\) is a odd number . That means \({ n \choose 1}, { n \choose 2}, { n \choose 3} .... { n \choose \frac {n-1}{2}} \) contains an odd numbers of odd numbers
So, \({ n \choose 1} + { n \choose 2}+ ...... +{ n \choose \frac {n-1}{2}} = \frac {1}{2} \times [{ n \choose 1} + { n \choose 2}+ ...... + { n \choose n-1}] = \frac {1}{2} \times ( 2^n - 2 ) = 2^{n-1} - 1\)
Where, \(2^{n-1} -1\) is a odd number . That means \({ n \choose 1}, { n \choose 2}, { n \choose 3} .... { n \choose \frac {n-1}{2}} \) contains an odd numbers of odd numbers
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