What does it mean?
Hey
hello again
What's the r'th term in the expansion of (a+b)^n ?
It is given by nCr*a^(n-r)*b^r . Right?
Yes, what's the r'th term in the expansion of (1+x)^n ?
It would be nCn*1^(n-n)*x^n ( nCn and 1^0 is 1 so it would be x^n).
Careful, I'm asking about the r'th term
Oops! nCr*x^r
Yes, r'th term in the expansion of (1+x)^n is nCr*x^r. What must we replace "r" by to see x^(n-2) ?
n-2
Yes, what's the "n-2" th term in the expansion of (1+x)^n ?
nC(n-2)*x^(n-2)
Right, so the coefficient of x^(n-2), according to our formula of r'th term is nC(n-2)
Oh sorry, the exponent is 2n, not n. Let's start over...
then It would be 2nC(n-2)x^(n-2)
Perfect! r'th term in the expansion of (1+x)^(2n) is (2n)Cr x^r we get x^(n-2) term if we replace r by n-2 above.
therefore we want to find the n-2 th term in the expansion of (1+x)^(2n) : (2n)C(n-2)*x^(n-2)
With me so far ?
Yes
simple expand (2n)C(n-2) and we're done
use the formula nCr = n!/[(n-r)! r!]
Oh! But in book they have expanded 2nC(n+2).
it's the same thing nCr = nC(n-r)
Ahh right, it must be some printing mistake. Replace 2nC(n+2) by 2nC(n-2) and everything should make sense...
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ncr = nc(n-r) but we don't want to use it here as rth term and n-r th terms wil have different degrees.
Also, why did they focused in x^(n-2)?... They could have chosen x^(n-1).
Good question. This will be answered after we look at the right hand side of the equation iii. Your goal here is to find an expression for C0C2 + C1C3 + C2C4 + ...., right ?
what is the full question?
Got it!! Thanks everyone..
You will be using this trick a lot in "generating functions" which you may be messing with soon. You get to find the coefficient of a certain term like, x^r, which represents number of ways of doing a particular thing. So it is okay if you don't get 100% of the problem now. You're gonna see more of these later and you will understand them after using this trick in different situations.
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