To prove by induction that

$\sum_{k=0}^n \left( 2^{n-k} \cdot \binom{n+k}{k} \right) = 4^n,$

we’ll proceed with two steps: Base Case and Inductive Step.

Step 1: Base Case ($n = 0$)

When $n = 0$, the sum has only one term:

$\sum_{k=0}^0 \left( 2^{0-k} \cdot \binom{0+k}{k} \right) = 2^0 \cdot \binom{0}{0} = 1.$

On the right-hand side, we have $4^0 = 1$. Therefore, the base case holds since both sides are equal.

Step 2: Inductive Step

Assume that for some integer $n = m$, the formula holds, i.e.,

$\sum_{k=0}^m \left( 2^{m-k} \cdot \binom{m+k}{k} \right) = 4^m.$

We need to show that

$\sum_{k=0}^{m+1} \left( 2^{(m+1)-k} \cdot \binom{(m+1)+k}{k} \right) = 4^{m+1}.$

To proceed, we split this sum at $m+1$:

$\sum_{k=0}^{m+1} \left( 2^{(m+1)-k} \cdot \binom{(m+1)+k}{k} \right) = \sum_{k=0}^m \left( 2^{(m+1)-k} \cdot \binom{(m+1)+k}{k} \right) + 2^0 \cdot \binom{(m+1) + (m+1)}{m+1}.$

Using the binomial properties and the inductive hypothesis here requires further decomposition,