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

The Correct Option is C

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

Answer 2

Step 1: Expand the series 

The given series is:

\[ S = \binom{30}{0} + 2 \cdot \binom{30}{1} \cdot 2 + 3 \cdot \binom{30}{2} \cdot 2 + \dots + 30 \cdot \binom{30}{30} \cdot 2. \]

Each term can be written as:

\[ n \left( \binom{30}{n} \right) 2. \] 

Step 2: Use the identity for weighted sums

The identity for such sums is:

\[ \sum_{k=0}^{n} k \cdot \left( \binom{n}{k} \right)^2 = n \cdot \binom{2n-1}{n-1}. \]

Substitute \( n = 30 \):

\[ S = 30 \cdot \binom{59}{29}. \] 

Step 3: Express \( \binom{59}{29} \) in factorials

Using the formula for combinations:

\[ \binom{59}{29} = \frac{59!}{29! \cdot 30!}. \]

Thus:

\[ S = 30 \cdot \frac{59!}{29! \cdot 30!}. \] 

Step 4: Compare with the given expression

The series is given as:

\[ S = \alpha \cdot \frac{60!}{(30!)^2}. \]

Substitute \( 60! = 60 \cdot 59! \):

\[ S = \alpha \cdot \frac{60 \cdot 59!}{(30!)^2}. \] 

Equating the two expressions \[ 30 \cdot \frac{59!}{29! \cdot 30!} = \alpha \cdot \frac{60 \cdot 59!}{(30!)^2}. \] Simplify: \[ 30 \cdot 29! \cdot 30 = \alpha \cdot 60. \] \[ \alpha = \frac{30 \cdot 30}{60} = 15. \]