$\sum_{k=0}^n\binom nk x^k=\sum_{k=0}^n\binom nk x^{n-k}.$

Question

$$\sum_{k=0}^n\binom nk x^k=\sum_{k=0}^n\binom nk x^{n-k}.$$

I want a deeper understanding of solving problems in this nature. I can´t grasp this writing way.I get confused just by looking at these problems. Please be patient with me. ;)

Answer 1

Hint

$${n\choose k}={n\choose n-k}$$

Answer 2

The sum on right hand side (RHS) of the equation is the sum on the left hand side (LHS) of the equation written backwards. Let me explain:

By definition of the sum, $$ \sum_{k=0}^n\binom nk x^k = \binom n0 x^0 + \binom n1 x^1 \ \ + \ \ ... \ \ + \binom {n}{n-1} x^{n-1}\ + \binom {n}{n} x^{n} \\ $$

writing the sum in reverse order to yeild, $$ = \binom {n}{n} x^{n} + \binom {n}{n-1} x^{n-1} \ \ + \ \ ... \ \ + \binom n1 x^1 +\binom n0 x^0\\ $$

with a bit more work we have,

$$ = \binom {n}{n-0} x^{n-0} + \binom {n}{n-1} x^{n-1}\ \ + \ \ ... \ \ + \binom n{n-(n-1)} x^{n-(n-1)} +\binom n{n-n} x^{n-n}\\ $$

that is, $$ \sum_{k=0}^n\binom nk x^k=\sum_{k=0}^n\binom n{n-k} x^{n-k}. $$

Finally we note the following combinatorial identity, $$ \begin{align} & \binom nk = \frac{n!}{k!(n-k)!} = \frac{n!}{(n-(n-k))!(n-k)!} = \binom n{n-k} \end{align} $$

Applying the combinatorial identity delivers the desired result, $$ \sum_{k=0}^n\binom nk x^k=\sum_{k=0}^n\binom nk x^{n-k}. $$