To solve the given problem:

Problem:

The sequence $(U_n)$ is defined by $U_n = n \cdot 2^n$ for all natural numbers $n$. We need to demonstrate that: $U_{n+2} = 4 \left(U_{n+1} - U_n\right)$

Step-by-Step Solution:

1. Express $U_{n+2}$, $U_{n+1}$, and $U_n$: Using the given formula $U_n = n \cdot 2^n$: $U_{n+2} = (n+2) \cdot 2^{n+2}$ $U_{n+1} = (n+1) \cdot 2^{n+1}$ $U_n = n \cdot 2^n$

2. Simplify $4 \left(U_{n+1} - U_n\right)$: First, calculate $U_{n+1} - U_n$: $U_{n+1} - U_n = \left((n+1) \cdot 2^{n+1}\right) - \left(n \cdot 2^n\right)$ Factor $2^n$ from both terms: $U_{n+1} - U_n = 2^n \left((n+1) \cdot 2 - n\right)$ Simplify: $U_{n+1} - U_n = 2^n \left(2n + 2 - n\right) = 2^n \left(n + 2\right)$

   Now, multiply by 4: $4 \left(U_{n+1} - U_n\right) = 4 \cdot 2^n \cdot (n+2)$ Simplify further: $4 \left(U_{n+1} - U_n\right) = (n+2) \cdot 2^{n+2}$

3. Compare $U_{n+2}$ and $4 \left(U_{n+1} - U_n\right)$: From the expressions above: $U_{n+2} = (n+2) \cdot 2^{n+2}$ and: $4 \left(U_{n+1} - U_n\right) = (n+2) \cdot 2^{n+2}$

   Since both expressions are identical, the equality: $U_{n+2} = 4 \left(U_{n+1} - U_n\right)$ is proven.

Conclusion:

The relation $U_{n+2} = 4 \left(U_{n+1} - U_n\right)$ holds for all natural numbers $n$.

Would you like further explanation on any of the steps? Here are some related questions you might find helpful:

1. How can the concept of recurrence relations be generalized for other sequences?
2. Why does factoring $2^n$ simplify the calculation?
3. What are the practical applications of such recursive sequences in real life?
4. How could you compute terms of this sequence efficiently using programming?
5. Can this approach be extended to non-linear recurrence relations?

Tip: Always check the first few terms of a recurrence relation to validate your results.