If $\sum_{\mathrm{r}=1}^{30} \frac{\mathrm{r}^{2}\left({ }^{30} \mathrm{C}_{\mathrm{r}}\right)^{2}}{{ }^{30} \mathrm{C}_{\mathrm{r}-1}}=\alpha \times 2^{29}$, then $\alpha$ is equal to ____ .

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Binomial Theorem

\sum_{r = 1}^{30} \frac{r ^{2} ( ^{30} C _{r} ) ^{2}}{^{30} C _{r - 1}} = α \times 2^{29}

, then

α

is equal to
____ .

Answer: 465

Using

\frac{^{30} C _{r}}{^{30} C _{r - 1}} = \frac{31 - r}{r}

\frac{r ^{2} ( ^{30} C _{r} ) ^{2}}{^{30} C _{r - 1}} = r^{2} \cdot^{30} C_{r} \cdot \frac{31 - r}{r} = r (31 - r)^{30} C_{r} .

r = 1 \sum 30 r (31 - r)^{30} C_{r} = 31 r = 1 \sum 30 r^{30} C_{r} - r = 1 \sum 30 r^{2}^{30} C_{r} .

Using

\sum r \cdot^{n} C_{r} = n \cdot 2^{n - 1}

and

\sum r^{2} \cdot^{n} C_{r} = n (n + 1) 2^{n - 2}

with

n = 30

31 (30 \cdot 2^{29}) - 30 \cdot 31 \cdot 2^{28} = 30 \cdot 31 \cdot 2^{28} (2 - 1) = 30 \cdot 31 \cdot 2^{28} .

Writing this as

α \cdot 2^{29}

30 \cdot 31 \cdot 2^{28} = \frac{30 \cdot 31}{2} \cdot 2^{29} = 465 \cdot 2^{29} .

α = 465.

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Integer answer