Question

I need an explanation for these steps.. Please check out the attachment.

Answer 1

Your multiplying itself times itself so any piece on the first should equal the same on the other.

Answer 2

But why did he chose x^(n-2)?

Answer 3

Ok so you get the thing they are saying about 2x^3+4x=ax^2+bx means it must be that a=2 and b=4?

Answer 4

And you get the n choose k part on (1+x)^2n?

If you just want to know why they choose that tern, I need to see more of what you are doing.

Answer 5

It's a problem related to binomial coefficients.
The actual question is:
If (1+x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ........ + C(n,n)x^n , Prove that :
C(n.0)C(n,2) + C(n,1)C(n,3) + C(n,2)C(n,4) + ........ + C(n,n-2)C(n,n) = 2n!/(n-2)!(n+2)!

Answer 6

ah this is induction then

Answer 7

So they will assume it is true for \(n-1\) and the further that to get to it being true for \(n\), i.e they assume the first x^{n-1} coefficients are the same, and use that to show all the coefficients are the same.

Answer 8

So they will assume it is true for n−1 and the further that to get to it being true for n, i.e they assume the first {n-1} coefficients are the same, and use that to show all n coefficients are the same.

Answer 9

Your first "why?" is answered by the binomial theorem
http://www.regentsprep.org/Regents/math/algtrig/ATP4/bintheorem.htm

if you have
\( (a+b)^n \), the coefficient for the \(b^m\) term is nCm (n choose m)
In your problem
\( (1+x)^{2n}\), the coefficient for the \(x^{n-2}\) terms is 2n Choose n-2
which is calculated as
\[ (2n)C(n-2)= \frac{(2n)!}{(n-2)!(2n-(n-2))!}= \frac{(2n)!}{(n-2)!(n+2)!} \]
By symmetry this also equals 2n Choose n+2.