Math combinatorics

Question

1. Consider the Taylor series 1 /(√(1 − 4x)) = a0 + a1x + a2x^2 + · · · which is valid in the range |x| < 1/4. Find a formula for an.
2. Give a combinatorial explanation for why 4^n = n∑ m=0 (2m choose m )(2(n − m) choose n − m ). What does this have to do with part 1?

Answer 1

1. If you expand it by Taylor series, you get

1 /(√(1 − 4x)) = 1+2x+6x^2+20x^3+70x^4+252x^5+924 x^6+3432x^7+12870x^8+48620 x^9+184756x^10+...

Now from here we can see that a_n=(2^n**2n-1)!!/n!

Here we used Taylor expansion as

f(x)=f(0)+f'(0)x+f''(0)x^2/2!+...

2 Use Binomial Expansion:

(a+b) ^n= a^n+n*a*b^(n-1)+...+nab^(n-1)+b^n.

Now set n=-1/2, a=1, b=-4x. Try to simplify

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