Prove using induction that \sum_{k=0}^n (2^{n-k} * ((n+k) \choose k)) is equal to 4^n

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,

Explore a step-by-step proof using induction to demonstrate that the summation \sum_{k=0}^n \left( 2^{n-k} \cdot \binom{n+k}{k} \right) equals 4^n. This problem covers induction, combinatorics, and exponential functions, suitable for high school students in Grades 11-12.

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