MCQ
If $\left({ }^{30} C _1\right)^2+2\left({ }^{30} C _2\right)^2+3\left({ }^{30} C _3\right)^2+\ldots \ldots+30\left({ }^{30} C _{30}\right)^2=$ $\frac{\alpha 60 !}{(30 !)^2}$, then $\alpha$ is equal to
- A$30$
- B$60$
- ✓$15$
- D$10$
$ S =30 \cdot(^{30} C _0)^2+29 \cdot{ }^{30} C _1)^2+28 \cdot{ }^{30} C _2)^2$
$+\ldots \ldots+0 \cdot{ }^{30} C _0)^2$
$\left.2 S =30 \cdot{ }^{30} C _0^2++^{30} C _1^2+\ldots \ldots \cdot+\cdot{ }^{30} C _{30}{ }^2\right)$
$S =15 \cdot{ }^{60} C _{30}=15 \cdot \frac{60 !}{(30 !)^2}$
$\frac{15 \cdot 10 !}{(30 !)^2}=\frac{\alpha \cdot 60 !}{(30 !)^2}$
$\Rightarrow \alpha=15$
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