1. **Stating the problem:** We want to verify or understand the formula for the sum $$\sum_{i=1}^{n+1} i(2^i) = n \cdot 2^{n+2} + 2$$. 2. **Recall the formula for the sum of the form** $$\sum_{i=1}^m i r^i$$ where $r \neq 1$: $$\sum_{i=1}^m i r^i = r \frac{1 - (m+1)r^m + m r^{m+1}}{(1-r)^2}$$ 3. **Apply this formula with** $r=2$ and $m = n+1$: $$\sum_{i=1}^{n+1} i 2^i = 2 \frac{1 - (n+2) 2^{n+1} + (n+1) 2^{n+2}}{(1-2)^2}$$ 4. **Simplify the denominator:** $$(1-2)^2 = (-1)^2 = 1$$ 5. **Simplify the numerator:** $$1 - (n+2) 2^{n+1} + (n+1) 2^{n+2} = 1 - (n+2) 2^{n+1} + (n+1) 2 \cdot 2^{n+1} = 1 - (n+2) 2^{n+1} + 2(n+1) 2^{n+1}$$ 6. **Combine like terms:** $$1 + \big(2(n+1) - (n+2)\big) 2^{n+1} = 1 + (2n + 2 - n - 2) 2^{n+1} = 1 + n 2^{n+1}$$ 7. **Multiply by 2:** $$2 \times \big(1 + n 2^{n+1}\big) = 2 + n 2^{n+2}$$ 8. **Rewrite the sum:** $$\sum_{i=1}^{n+1} i 2^i = n 2^{n+2} + 2$$ **Final answer:** The given formula is correct. This shows the sum $$\sum_{i=1}^{n+1} i(2^i)$$ equals $$n \cdot 2^{n+2} + 2$$ as stated.