If , then is equal to __________. [2025]

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Solution

Q.
If $\sum_{r = 1}^{30} \frac{r^{2} {({}^{30}C_{r})}^{2}}{{}^{30}C_{r - 1}} = α \times 2^{29}$ , then $α$ is equal to __________. [2025]

Ans.
(465)

$\sum_{r = 1}^{30} \frac{r^{2} {({}^{30}C_{r})}^{2}}{{}^{30}C_{r - 1}} = \sum_{r = 1}^{30} r^{2} (\frac{31 - r}{r}) \cdot \frac{30!}{r! (30 - r)!}$

$= \sum_{r = 1}^{30} \frac{(31 - r) 30!}{(r - 1)! (30 - r)!} = 30 \sum_{r = 1}^{30} (30 - r + 1) \times {}^{29}C_{30 - r}$

$= 30 (\sum_{r = 1}^{30} (30 - r) \times {}^{29}C_{30 - r} + \sum_{r = 1}^{30} {}^{29}C_{30 - r})$

$= 30 (29 \times 2^{28} + 2^{29})$

$[∵ {}^{n}C_{1} + 2 {}^{n}C_{2} + 3 {}^{n}C_{3} + 4 {}^{n}C_{4} + . . . + n {}^{n}C_{n} = n 2^{n - 1} and {}^{n}C_{1} + {}^{n}C_{2} + {}^{n}C_{3} + . . . + {}^{n}C_{n} = 2^{n - 1}]$

$= 30 (29 + 2) 2^{28} = 15 \times 31 \times 2^{29} = 465 (2^{29})$

$∴ α = 465$ .

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