Binomial Theorem
Combinatorial Identity / Sum
nta_pyq_2025_apr
Grade 11
Question:
If $\displaystyle\sum_{r=1}^{30} \frac{r^2 \left({}^{30}C_r\right)^2}{{}^{30}C_{r-1}} = \alpha \times 2^{29}$, then $\alpha$ is equal to ___
Step-by-Step Solution
Key Concept: Simplify ${}^{30}C_r / {}^{30}C_{r-1} = (31-r)/r$ and transform the summation into manageable standard sums using ${}^{29}C_{30-r}$.
$\sum_{r=1}^{30} \frac{r^2 ({}^{30}C_r)^2}{{}^{30}C_{r-1}} = \sum_{r=1}^{30} r^2 \cdot \frac{31-r}{r} \cdot \frac{30!}{r!(30-r)!} = 30\sum_{r=1}^{30}(30-r+1){}^{29}C_{30-r} = 30(29 \times 2^{28} + 2^{29}) = 30 \times 31 \times 2^{28}/2 = 465 \times 2^{29}$. So $\alpha = 465$.
Correct Answer: 465