1) show that
a) 2(2n-1 C n-1) = (2n Cn)
b) n(2n C n) = (n+1)(2n C n+1)
To prove the given equations, we will use the property of binomial coefficients - Pascal's identity. Pascal's identity states that for any positive integers n and k, the binomial coefficient of n and k can be expressed using the binomial coefficient of n-1 and k-1, and the binomial coefficient of n-1 and k.
a) 2(2n-1 C n-1) = (2n C n)
To prove this equation, let's start by expanding the binomial coefficients on both sides of the equation.
First, let's expand 2(2n-1 C n-1):
2(2n-1 C n-1) = 2 * [(2n-1)! / (n-1)!(2n-1-(n-1))!]
Simplifying this:
2(2n-1 C n-1) = 2 * [(2n-1)! / (n-1)!(n)!(2n-1-n)!(n-1)!]
= 2 * [(2n-1)! / (n)!(n-1)!(2n-2)!(n-1)!]
= 2 * [(2n-1)! / (n)!(n-1)!(2n-2)!]
Next, let's expand (2n C n):
(2n C n) = [(2n)! / (n!(2n-n)!)]
Simplifying this:
(2n C n) = [(2n)! / (n)!(n)!(2n-2)!]
Since (2n-2)! is common in both expressions, let's cancel it out:
2(2n-1 C n-1) = 2 * [(2n-1)! / (n)!(n-1)!(2n-2)!]
= [(2n)! / (n)!(n)!(2n-2)!]
By canceling out common terms, we can see that 2(2n-1 C n-1) = (2n C n).
b) n(2n C n) = (n+1)(2n C n+1)
To prove this equation, let's expand both sides of the equation using binomial coefficients.
First, let's expand n(2n C n):
n(2n C n) = n * [(2n)! / (n!(2n-n)!)]
Simplifying this:
n(2n C n) = n * [(2n)! / (n)!(n)!(2n-2)!]
= [(2n)! / (n)!(n)!(2n-2)!] * n
Now, let's expand (n+1)(2n C n+1):
(n+1)(2n C n+1) = (n+1) * [(2n)! / (n+1)!(2n-(n+1))!]
Simplifying this:
(n+1)(2n C n+1) = (n+1) * [(2n)! / (n+1)!(n)!(2n-n-1)!(n+1)!]
= [(2n)! / (n)!(n+1)!(2n-1-n-1)!] * (n+1)
= [(2n)! / (n)!(n+1)!(2n-2)!(n+1)!] * (n+1)
By canceling out common terms, we can see that n(2n C n) = (n+1)(2n C n+1).
Thus, we have proved both equations.