Answer: Using factorials,
\(\binom{n}{r-1}+\binom{n}{r}=\frac{n!}{(r-1)!(n-r+1)!}+\frac{n!}{r!(n-r)!}\).
Now take a common factor of \(\frac{n!}{(r-1)!(n-r)!}\):
\(\binom{n}{r-1}+\binom{n}{r}=\frac{n!}{(r-1)!(n-r)!}\left(\frac{1}{n-r+1}+\frac{1}{r}\right)\).
Combine the terms in brackets:
\(\frac{1}{n-r+1}+\frac{1}{r}=\frac{r+(n-r+1)}{r(n-r+1)}=\frac{n+1}{r(n-r+1)}\).
So
\(\binom{n}{r-1}+\binom{n}{r}=\frac{n!}{(r-1)!(n-r)!}\cdot \frac{n+1}{r(n-r+1)}\).
Since \(r! = r(r-1)!\) and \((n-r+1)!=(n-r+1)(n-r)!\), this becomes
\(\binom{n}{r-1}+\binom{n}{r}=\frac{(n+1)!}{r!(n-r+1)!}=\binom{n+1}{r}\).
Hence, for the binomial expansion, define
\(H_n:\ (a+x)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}x+\cdots+\binom{n}{r}a^{n-r}x^r+\cdots+\binom{n}{n}x^n\).
We prove \(H_n\) by induction.
First, when \(n=1\),
\((a+x)^1=a+x=\binom{1}{0}a+\binom{1}{1}x\),
so \(H_1\) is true.
Assume \(H_k\) is true for some positive integer \(k\). Then
\((a+x)^k=\binom{k}{0}a^k+\binom{k}{1}a^{k-1}x+\cdots+\binom{k}{r}a^{k-r}x^r+\cdots+\binom{k}{k}x^k\).
Multiply both sides by \((a+x)\):
\((a+x)^{k+1}=(a+x)\sum_{r=0}^{k}\binom{k}{r}a^{k-r}x^r\).
When we collect the coefficient of \(a^{k-r+1}x^r\), it comes from two terms: \(a\cdot \binom{k}{r}a^{k-r}x^r\) and \(x\cdot \binom{k}{r-1}a^{k-r+1}x^{r-1}\). So the coefficient is
\(\binom{k}{r-1}+\binom{k}{r}=\binom{k+1}{r}\).
Therefore
\((a+x)^{k+1}=\binom{k+1}{0}a^{k+1}+\binom{k+1}{1}a^kx+\cdots+\binom{k+1}{r}a^{k+1-r}x^r+\cdots+\binom{k+1}{k+1}x^{k+1}\),
so \(H_{k+1}\) is true.
Since \(H_1\) is true and \(H_k \Rightarrow H_{k+1}\), it follows by mathematical induction that the result holds for every positive integer \(n\).
We first establish the identity for consecutive binomial coefficients.
Start from the factorial definitions:
\(\binom{n}{r-1}=\frac{n!}{(r-1)!(n-r+1)!}\), \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\).
Hence
\(\binom{n}{r-1}+\binom{n}{r}=\frac{n!}{(r-1)!(n-r+1)!}+\frac{n!}{r!(n-r)!}\).
Rewrite the second term so both terms have a common factor \(\frac{n!}{(r-1)!(n-r)!}\):
\(\binom{n}{r-1}+\binom{n}{r}=\frac{n!}{(r-1)!(n-r)!}\left(\frac{1}{n-r+1}+\frac{1}{r}\right)\).
Now simplify the bracket:
\(\frac{1}{n-r+1}+\frac{1}{r}=\frac{r+(n-r+1)}{r(n-r+1)}=\frac{n+1}{r(n-r+1)}\).
So
\(\binom{n}{r-1}+\binom{n}{r}=\frac{n!}{(r-1)!(n-r)!}\cdot \frac{n+1}{r(n-r+1)}\).
Using \(r(r-1)!=r!\) and \((n-r+1)(n-r)!=(n-r+1)!\), this becomes
\(\binom{n}{r-1}+\binom{n}{r}=\frac{(n+1)!}{r!(n-r+1)!}=\binom{n+1}{r}\).
This is the required identity.
Now prove the binomial expansion by induction on \(n\).
Let
\(H_n:\ (a+x)^n=\binom{n}{0}a^n+\binom{n}{1}a^{n-1}x+\cdots+\binom{n}{r}a^{n-r}x^r+\cdots+\binom{n}{n}x^n\).
For \(n=1\),
\((a+x)^1=a+x=\binom{1}{0}a+\binom{1}{1}x\),
so \(H_1\) is true.
Assume \(H_k\) is true for some positive integer \(k\). Then
\((a+x)^k=\sum_{r=0}^{k}\binom{k}{r}a^{k-r}x^r\).
Multiply by \((a+x)\):
\((a+x)^{k+1}=(a+x)\sum_{r=0}^{k}\binom{k}{r}a^{k-r}x^r\).
On expanding, the coefficient of \(a^{k+1-r}x^r\) in \((a+x)^{k+1}\) comes from two parts: choosing \(a\) from the factor \((a+x)\) and choosing \(x\) from the factor \((a+x)\). Therefore the coefficient is
\(\binom{k}{r}+\binom{k}{r-1}\).
By the identity proved above, this equals \(\binom{k+1}{r}\). Hence
\((a+x)^{k+1}=\binom{k+1}{0}a^{k+1}+\binom{k+1}{1}a^kx+\cdots+\binom{k+1}{r}a^{k+1-r}x^r+\cdots+\binom{k+1}{k+1}x^{k+1}\).
So \(H_{k+1}\) is true whenever \(H_k\) is true.
Since \(H_1\) is true and \(H_k\Rightarrow H_{k+1}\), the result follows by mathematical induction for every positive integer \(n\).
