To prove the given summation formula:

$\sum_{k=0}^h k \cdot 2^k = (h-1) \cdot 2^{h+1} + 2,$

we proceed with mathematical induction.

---

Step 1: Base Case $h = 0$

The left-hand side (LHS) becomes:

$\sum_{k=0}^0 k \cdot 2^k = 0 \cdot 2^0 = 0.$

The right-hand side (RHS) is:

$(h-1) \cdot 2^{h+1} + 2 = (0-1) \cdot 2^{0+1} + 2 = (-1) \cdot 2^1 + 2 = -2 + 2 = 0.$

Thus, LHS = RHS for $h = 0$. The base case holds true.

---

Step 2: Induction Hypothesis

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

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

---

Step 3: Induction Step

We need to prove that the formula holds for $h = n+1$, i.e.,

$\sum_{k=0}^{n+1} k \cdot 2^k = \big((n+1)-1\big) \cdot 2^{(n+1)+1} + 2.$

Simplify the RHS:

$\big((n+1)-1\big) \cdot 2^{(n+1)+1} + 2 = n \cdot 2^{n+2} + 2.$

Now, consider the left-hand side:

$\sum_{k=0}^{n+1} k \cdot 2^k = \sum_{k=0}^n k \cdot 2^k + (n+1) \cdot 2^{n+1}.$

From the induction hypothesis, we know:

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

Substitute this into the LHS:

$\sum_{k=0}^{n+1} k \cdot 2^k = \big((n-1) \cdot 2^{n+1} + 2\big) + (n+1) \cdot 2^{n+1}.$

Combine like terms:

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

Factor out $2^{n+1}$:

$\sum_{k=0}^{n+1} k \cdot 2^k = \big((n-1) + (n+1)\big) \cdot 2^{n+1} + 2.$

Simplify $(n-1) + (n+1)$:

$(n-1) + (n+1) = 2n.$

Thus:

$\sum_{k=0}^{n+1} k \cdot 2^k = 2n \cdot 2^{n+1} + 2.$

Simplify the first term:

$2n \cdot 2^{n+1} = n \cdot 2^{n+2}.$

Therefore:

$\sum_{k=0}^{n+1} k \cdot 2^k = n \cdot 2^{n+2} + 2.$

This matches the RHS for $h = n+1$, completing the induction step.

---

Conclusion

By the principle of mathematical induction, the formula holds for all $h \geq 0$:

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

---

Do you want me to clarify any step further or expand on any part of the proof?

---

Related Questions

1. How do you prove summation identities using mathematical induction?
2. Can you derive a general formula for $\sum_{k=0}^h k^2 \cdot 2^k$?
3. What happens if the base changes from $2$ to a general $b$?
4. How can you use generating functions to derive such formulas?
5. What are other approaches to prove summation formulas (e.g., telescoping)?

Tip: Mathematical induction works best when you start small, assume for $n$, and prove for $n+1$. Always simplify step by step.