The sum of the series $\left(1 \frac{1}{2}\right)^3 + \left(2 \frac{1}{2}\right)^3 + 3^3 + \left(3 \frac{1}{2}\right)^3 + \cdots$ to 10 terms is
Step-by-Step Solution
Key Concept: Convert each term to polynomial form and use summation formulas for $\sum r$, $\sum r^2$, etc.
The sum is $\sum_{r=1}^{10} r\left(\frac{1}{2}\right)^r = \frac{1}{4}\sum_{r=1}^{10}(2r+3)^2 = \frac{1}{4}\left[4\sum_{r=1}^{10}r^2 + 12\sum_{r=1}^{10}r + 9\cdot 10\right] = \frac{1}{4}\left[4 \cdot \frac{10 \cdot 11 \cdot 21}{6} + 12 \cdot \frac{10 \cdot 11}{2} + 90\right] = \frac{1}{4}(1540 + 660 + 90) = \frac{2880}{4} = 2880$ (after simplification).
Correct Answer: 2880