7.Binomial Theorem
hard

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

$60$

C

$15$

D

$10$

(JEE MAIN-2023)

Solution

$S =0 .\left({ }^{30} C _0\right)^2+1 \cdot\left(\cdot{ }^{30} C _1\right)^2+2 \cdot\left({ }^{30} C _2\right)^2+\ldots \ldots+30 \cdot\left({ }^{30} C _{30}\right)^2$

$ S =30 \cdot(^{30} C _0)^2+29 \cdot{ }^{30} C _1)^2+28 \cdot{ }^{30} C _2)^2$

$+\ldots \ldots+0 \cdot{ }^{30} C _0)^2$

$\left.2 S =30 \cdot{ }^{30} C _0^2++^{30} C _1^2+\ldots \ldots \cdot+\cdot{ }^{30} C _{30}{ }^2\right)$

$S =15 \cdot{ }^{60} C _{30}=15 \cdot \frac{60 !}{(30 !)^2}$

$\frac{15 \cdot 10 !}{(30 !)^2}=\frac{\alpha \cdot 60 !}{(30 !)^2}$

$\Rightarrow \alpha=15$

Standard 11
Mathematics

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