91Ó°ÊÓ

In this step, we want to show that the binomial theorem holds for n=0. According to the theorem, \[(a+b)^0=\sum_{k=0}^0 {0 \choose k} a^{0-k}b^k\] Now, we know that any non-zero number to the power of 0 is equal to 1, so \((a+b)^0 = 1\). As for the right-hand side of the equation, there's only one term in the summation since k=0 is the only value in the range, so we have: \[\sum_{k=0}^0 {0 \choose k} a^{0-k}b^k = {0 \choose 0}a^0b^0\] By definition, \({n \choose n} = {0 \choose 0} = 1\). Also, \(a^0 = b^0 = 1\). Therefore, \[{0 \choose 0}a^0b^0 = 1 * 1 * 1 = 1\] Since both the left-hand side and the right-hand side of the equation equal 1, we have proved the base case: \[(a+b)^0=\sum_{k=0}^0 {0 \choose k} a^{0-k}b^k\]

In this step, we want to assume the binomial theorem is true for n=p and show that it also holds for n=p+1. 1. Inductive Hypothesis: Assume the theorem is true for n=p, i.e., \[(a+b)^p = \sum_{k=0}^p {p \choose k} a^{p-k} b^k\] 2. Now, let's show that it also holds for n=p+1, \[(a+b)^{p+1} = \sum_{k=0}^{p+1} {p+1 \choose k} a^{p+1-k} b^k\] To do this, we'll start by expanding \((a+b)^{p+1}\). \[(a+b)^{p+1} = (a+b)(a+b)^p\] Now, using the inductive hypothesis, we can substitute the expression on the right-hand side of the equation: \[(a+b)^{p+1} = (a+b)\left(\sum_{k=0}^p {p \choose k} a^{p-k} b^k\right)\] Next, we distribute the (a+b) term: \[(a+b)^{p+1} = \sum_{k=0}^p {p \choose k} a^{p+1-k} b^k + \sum_{k=0}^p {p \choose k} a^{p-k} b^{k+1}\] Now, we rewrite the second summation with the index shifted by 1 (replace k with k-1): \[(a+b)^{p+1} = \sum_{k=0}^p {p \choose k} a^{p+1-k} b^k + \sum_{k=1}^{p+1} {p \choose k-1} a^{p+1-k} b^{k}\] We can combine the two summations into one: \[(a+b)^{p+1} = \sum_{k=0}^{p+1} \left({p \choose k} a^{p+1-k} b^k + {p \choose k-1} a^{p+1-k} b^{k}\right)\] Now, we'll use the identity for the binomial coefficients: \[{n \choose k} + {n \choose k-1} = {n+1 \choose k}\] Applying this identity, we get: \[(a+b)^{p+1} = \sum_{k=0}^{p+1} {p+1 \choose k} a^{p+1-k} b^k\] This is what we wanted to prove; the binomial theorem holds for n=p+1. Since we have proved both the base case and the inductive step, we can conclude that the binomial theorem is true for all natural numbers n, using mathematical induction.