If`((n),(r )) "denotes " ""^nC_r then `
(a) Evalutae : `2^(15)((30),(0))((30),(1))-2^(14)((30),(1))((29),(14))+2^(13)((30),(2))((28),(13)).......-((30),(15))((15),(0))`
( b) Prove that : `Sigma_(r=1)^(n) ((n-1),(n-r))((n),(r))=((2n-1),(n-1))` ( c) Prove that : `((n),(r))((r),(k))=((n),(k))((n-r),(r-k))`

Answer

Let's solve the given problems step by step.

### Part (a)
We need to evaluate the expression:

\[ 2^{15} \binom{30}{0} \binom{30}{1} - 2^{14} \binom{30}{1} \binom{29}{14} + 2^{13} \binom{30}{2} \binom{28}{13} - \ldots - \binom{30}{15} \binom{15}{0} \]

This expression can be simplified using the binomial theorem. The general term in the series can be recognized as:

\[ (-1)^k 2^{15-k} \binom{30}{k} \binom{30-k}{15-k} \]

for \( k = 0, 1, 2, \ldots, 15 \).

This is equivalent to:

\[ \sum_{k=0}^{15} (-1)^k 2^{15-k} \binom{30}{k} \binom{30-k}{15-k} \]

By applying the binomial theorem, we can simplify this sum. The sum represents the coefficient of \( x^{15} \) in the expansion of:

\[ (1 - 2x)^{30} \]

Thus, we need to evaluate:

\[ (1 - 2)^{30} = (-1)^{30} = 1 \]

So, the final answer for part (a) is:

\[ \boxed{1} \]