If $$\binom{30}{1}^2 + 2\binom{30}{2}^2 + 3\binom{30}{3}^2 + \ldots + 30\binom{30}{30}^2 = \frac{\alpha \cdot 60!}{(30!)^2}$$, then $$\alpha$$ is equal to

Name: If \\binom{30}{1}^2 + 2\\binom{30}{2}^2 + 3\\binom{30}{3}^2 + \\ldots + 30\\binom{30}{30}^2 = \\frac{\\alpha \\cdot 60!}{(30!)^2}, then \\alpha is equal to
Uploaded: 2026-05-28T16:07:28.181764+05:30
Duration: 4 min 43 s
Description: If \\binom{30}{1}^2 + 2\\binom{30}{2}^2 + 3\\binom{30}{3}^2 + \\ldots + 30\\binom{30}{30}^2 = \\frac{\\alpha \\cdot 60!}{(30!)^2}, then \\alpha is equal to

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If $$\binom{30}{1}^2 + 2\binom{30}{2}^2 + 3\binom{30}{3}^2 + \ldots + 30\binom{30}{30}^2 = \frac{\alpha \cdot 60!}{(30!)^2}$$, then $$\alpha$$ is equal to

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