Binomial Theorem Question 3
Question 3 - 24 January - Shift 2
If $({ }^{30} C_1)^{2}+2((^{30} C_2)^{2}+3({ }^{30} C_3)^{2}+\ldots \ldots . .+30({ }^{30} C _{30})^{2}=.$ $\frac{\alpha 60 !}{(30 !)^{2}}$, then $\alpha$ is equal to
(1) 30
(2) 60
(3) 15
(4) 10
Show Answer
Answer: (3)
Solution:
Formula: Important Results on Binomial Coefficients (xii)
$S=0 .({ }^{30} C_0)^{2}+1 \cdot({ }^{30} C_1)^{2}+2 \cdot({ }^{30} C_2)^{2}+\ldots \ldots .+30 \cdot({ }^{30} C _{30})^{2}$
$S=30 .{ }^{30} C_0)^{2}+29 \cdot({ }^{30} C_1)^{2}+28 \cdot({ }^{30} C_2)^{2}+\ldots \ldots+0 .{ }^{30} C_0)^{2}$
$2 S=30 \cdot({ }^{30} C_0^{2}+{ }^{30} C_1{ }^{2}+\ldots \ldots .+{ }^{30} C _{30}{ }^{2})$
$S=15 \cdot{ }^{60} C _{30}=15 \cdot \frac{60 !}{(30 !)^{2}}$
$\frac{15 \cdot 10 !}{(30 !)^{2}}=\frac{\alpha \cdot 60 !}{(30 !)^{2}}$
$\Rightarrow \alpha=15$
