Specialization Math Example 1

Solution

1. 1
   The Binomial Theorem gives $(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$. Specialization means substituting specific values for the parameters to obtain concrete identities.
2. 2
   Substitute $a=1, b=1$: $(1+1)^n = \sum_{k=0}^{n}\binom{n}{k}1^{n-k}\cdot 1^k = \sum_{k=0}^{n}\binom{n}{k}$. Since the left side equals $2^n$, we get the identity $\sum_{k=0}^{n}\binom{n}{k} = 2^n$.
3. 3
   Substitute $a=1, b=-1$: $(1-1)^n = \sum_{k=0}^{n}\binom{n}{k}(-1)^k$. For $n \ge 1$ the left side is $0^n = 0$, giving the alternating-sum identity $\sum_{k=0}^{n}\binom{n}{k}(-1)^k = 0$.