Question

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

• A. 60
• B. 10
• C. 15
• D. 30

Answer 1

Answer: (C)

$S=0\cdot {\left({}^{30}{C}_0\right)}^2+1\cdot {\left({}^{30}{C}_1\right)}^2+2\cdot \cdot {\left({}^{30}{C}_2\right)}^2+\dots \dots +30\cdot {\left({}^{30}{C}_{30}\right)}^2$
$S=30.{\left({}^{30}{C}_0\right)}^2+29.{\left({}^{30}{C}_1\right)}^2+28.{\left({}^{30}{C}_2\right)}^2+\dots ..+0.{\left({}^{30}{C}_0\right)}^2$
$2S=30\cdot \left({}^{30}{C}_0{}^2+{+}^{30}{C}_1^2+\dots \dots .\cdot {}^{30}{C}_{30}{}^2\right)$
$S=15\cdot {}^{60}{C}_{30}=15\cdot \frac{60!}{(30!{)}^2}$
$\frac{15\cdot 10!}{(30!{)}^2}=\frac{\alpha \cdot 60!}{(30!{)}^2}$
$\Rightarrow \alpha =15$