How to proof the Pascal's identity using the identity $(a+b)^n=(a+b)^{n-1}a+(a+b)^{n-1}b$?

Question

Using the identity $(a+b)^n=(a+b)^{n-1}a+(a+b)^{n-1}b$ to prove the sum property of binomial coefficients:

$\binom{n}{k}=\binom{n-1}{r-1}+\binom{n-1}{k}.$

I don't get the idea of start to solve this. Can is suppose that the coeficient of $a^{n-k}b^k$ is $\binom{n}{k}$?

Answer 1

You're right, you need to use that $$ (a+b)^{n-1} = \sum_{k=0}^{n-1} \binom{n-1}{k}a^{n-1-k}b^k$$ and $$ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^k$$ Then using the identity you've been provided you need to prove that $$ \binom{n}{k} = \binom{n-1}{{\color{red} k}-1}+\binom{n-1}{k}$$

Answer 2

Well, its just a comparison of coefficients:

The coefficient of $a^kb^{n-k}$ in $(a+b)^n$ is $n\choose k$.

The coefficient of $a^{k-1}b^{n-1-(k-1)}a$ in $(a+b)^{n-1}a$ is $n-1\choose k-1$.

The coefficient of $a^kb^{n-1-k}b$ in $(a+b)^{n-1}b$ is $n-1\choose k$.

Now sum up.