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38 Show that al n onn egative integ ers n can be represented uniquely in the form n (3)+()+i where @, b, and integ er s with 0 a < b < c, (This is called the ...

Question

38 Show that al n onn egative integ ers n can be represented uniquely in the form n (3)+()+i where @, b, and integ er s with 0 a < b < c, (This is called the binomial number system:)

38 Show that al n onn egative integ ers n can be represented uniquely in the form n (3)+()+i where @, b, and integ er s with 0 a < b < c, (This is called the binomial number system:)



Answers

If $n$ is a positive integer, show that $$ \left(\begin{array}{l}{n} \\ {0}\end{array}\right)+\left(\begin{array}{l}{n} \\ {1}\end{array}\right)+\cdots+\left(\begin{array}{l}{n} \\ {n}\end{array}\right)=2^{n} $$ [Hint: $\left.2^{n}=(1+1)^{n} ; \text { now use the Binomial Theorem. }\right]$

Mhm. The binomial theorem states that for an expression X plus Y raised to end can be solved using binomial theorem as and C zero. All extra the bubble and plus n c one extra the bubble and minus one name's white dr dot plus n C N. Why do the power end? So a few things to note your our the total number of terms will be endless one and the power of each term would be. And so far the expression total power. And we can rewrite this as one plus one to the power N. Which can be rewritten as N C zero. Want to the power in plus n C. One Yeah. One to the powerball and minus one times one plus and C two onto the power and minus two times one square. Still N C n one to the power. And and we know that one raised to any power is simply one. So we can write that due to the power N is equal to n C zero plus n C one. Okay, up to and see. And

The formula for n choose J his n factorial over j factorial times n minus j factorial. So if we have n choose n minus one, that's equal in fact well over J factorial, which in this case it was n minus one. Factorial comes in minus and minus one factorial the smear in as in factorial, we're n minus one factorial times and minus and plus one factorial. The Sequels and factorial end minus one. Factorial comes one factory now one factorial just equals once we can disregard that term and we're left with n factorial over n minus one. Facts were you now in factorial equals n times n minus one comes in minus two, etcetera an end minus one Factorial He was n minus one comes end minus two. Except so you could see that these two terms would cancel out. These two terms would cancel on every term will cancel out except the end. So this whole thing is evil toe end. So we've just shown that n choose n minus one is equal to end Now if we have n choose n that's equal to n factorial over and factorial times en minus and factorial So the two in fact rules cancel out. And we're left with one over n minus end, factorial and minus and factorial. He goes zero factorial, which equals one. So so think equals 1/0 factorial, which equals 1/1, which was one. So we've just shown the end shoes and equals one.

Mhm N c r can be represented as N factorial upon. N minus R factorial times are in fact told you. So N c n minus one can be found as in factorial upon and minus of n minus one, I am n minus one factorial which would be equal to n factorial giants, N minus n plus one factorial times, n minus one factorial. So this term becomes simply one factory which is one. And n factorial can be written as N, multiplied by n minus one factorial upon n minus one factorial. So these to get cancer and this is equal to n similarly and n c n is equal to n factorial upon and minus and factorial times and factorial. So this becomes and factorial divided by zero factorial times and factorial zero factorial is equal to one. So this result is simply equal to

Okay, let's start first with a communitarian argument. So suppose that we have to and distinguishable elements and we wanna choose to without repetition. So we can do these in two and communitarian number with two waste. Okay, Now, let's come this in a different way. Suppose that we have the balls, the elements separating 22 works. We have the option off selected two elements from the first bath, and we can do that in in community aerial number two ways. And suppose we have the other option off selecting the two elements from the second. And we can do that the same way in in, um, communitarian number two ways. Now, what about if we select one element from one murder in the other elements from the other back? If we just one from the first body, we will have any options. And the 2nd 1 from the second bag don't give us another in motion. So we have in times. So in total, there will be in a communitarian number two plus in community and number two plus in square options. That is two times the communitarian number eight on two. Bless mind square possibilities. So we have that this way of counting to win community real number two and this way of counting that we just find out, are the same now. Algebraic Lee. It's slightly easier. So we have the communitarian number two in with two that is two and factorial over to factorial too and minus two factory. We can break these factorial into to end times two and minus one times two and minus two Factorial. Now we can simplify peace with this. Remember that two factorial is too. So we have to win times two and minus one day by the bike to that is Foreign Square minus two and divide by two. Now we can take one of these to off these and squares too and squares from here and we'll have to end. Square over to that is and square and we have left on top two and square minus two in now we come fact, arise one in in a tube from these both terms. So we have to win times in minus one. And now we're very close to what we wanted because we can multiply on the top and on the bottom, by and minus two and that would give us these three terms on top, combining give us the in fact Aurea. And on the bottom, we have two factorial in minus one fact Odio. And so this is a community aerial number and two, and that is the identity that we were section for.


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