Pure Maths---Binomial

a) nCr + n C(r-1) = n!/r!/(n-r)! + n!/(r-1)! / (n-r+1)!
=n!/ (r-1)! / (n-r)! [ 1/r + 1/(n-r+1)]
= n!/ (r-1)! / (n-r)! X (n+1)/r / (n-r+1)
= (n+1)!/ r! / (n+1-r)!
= n+1 C r

b)
f(n+1) = S (n+1+k) C (n+1) 1/2^k
= S [ (n+k) C (n+1) + (n+k) C n ] 1/2^k
=S (n+k) C (n+1) 1/ 2^k + S (n+k) C n 1/2^k
= S (n+k) C (n+1) 1/ 2^k + f(n)

Since n+k >= n+1 therefore, k>=1
however, the first term is sum up from k=0 to k=n
we separate it
(n+0) C (n+1) X 1/ 2^0 + S (n+k) C (n+1) 1/ 2^k ; the first term is k=0
= 1 + S (n+k) C (n+1) 1/ 2^k ; the first term equal to 1 by definition
= S (n+k) C (n+1) 1/ 2^k ; which is equal to the sum from k=0 to k=n
= f(n)

thus, f(n+1)=2 f(n)

for all n in N
f(0)= 1
f(1)= 2X1
f(2) = 2X f(1)= 2^2
thus, we can see,
f(n) = 2^n

you can proof by MI, but it is not necessary.