Question

If $\left({ }^{30} C _1\right)^2+2\left({ }^{30} C _2\right)^2+3\left({ }^{30} C _3\right)^2+\ldots+30\left({ }^{30} C _{30}\right)^2=\frac{\alpha 60 !}{(30 !)^2}$ then $\alpha$ is equal to :

• A. 60
• B. 10
• C. 15
• D. 30

Answer 1

Answer: (C)

15

Answer 2

The correct answer is (C) : 15
S=0⋅(30C0​)2+1⋅(30C1​)2+2⋅⋅(30C2​)2+……+30⋅(30C30​)2
S=30.(30C0​)2+29.(30C1​)2+28.(30C2​)2+…..+0.(30C0​)2
2S=30⋅(30C0​2++30C12​+…….⋅30C30​2)
S=15⋅60C30​=15⋅(30!)260!​
(30!)215⋅10!​=(30!)2α⋅60!​
⇒α=15