$^nC_r$ summation series

Question

Given, COROLLARY 3 in Kennith H Rosen Discrete Mathematics book,
$$\sum_{k=0}^{n} 2^{k} \binom{n}{k} = 3^{n}$$

Is this below mentioned result valid too? $$\sum_{k=0}^{n} 2^{n-k} \binom{n}{k} = 3^{n}$$

If not, then how to solve it?

Answer 1

Both statements are true.

Proof

By Binomial Theorem, we have

$$\sum_{k=0}^{n} \binom{n}{k} x^k = (x+1)^n $$

Let x=2, we have $$\sum_{k=0}^{n} 2^k\binom{n}{k} = 3^n \tag{1}\label{eq1}$$

Now define k' = n - k so that k = n - k'

It follows that $$\sum_{k'=0}^{n} 2^{n - k'}\binom{n}{k'} = 3^n$$ since k' is a dummy variable, we can call it k again without loss of generality to get $$ \sum_{k=0}^{n} 2^{n -k}\binom{n}{k} = 3^n \tag{2}\label{eq2}$$

The conclusion follows from equations (1) and (2).

QED