`((1 + ""^nC_1 + ""^nC_2 + ""^nC_3+…….+nC_n)^2)/(1 + ""^(2n)C_1 + ""^(2n)C_2 + ""^(2n)C_3 + ……… + ""^(2n)C_(2n)) = `

Options

A. 1
B. `-1`
C. 2
D. 3

Answer

To solve the given problem, we need to evaluate the expression:

\[ \frac{(1 + \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ldots + \binom{n}{n})^2}{(1 + \binom{2n}{1} + \binom{2n}{2} + \binom{2n}{3} + \ldots + \binom{2n}{2n})} \]

**Step 1: Simplifying the Numerator**

The numerator is \(1 + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n}\). We can use the Binomial Theorem, which states that:

\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

By substituting \(x = 1\), we get:

\[ (1 + 1)^n = 2^n = \sum_{k=0}^{n} \binom{n}{k} \]

Thus, we have:

\[ 1 + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n \]

Now, squaring this result for the numerator:

\[ (1 + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n})^2 = (2^n)^2 = 2^{2n} \]

**Step 2: Simplifying the Denominator**

Now, we simplify the denominator, which is:

\[ 1 + \binom{2n}{1} + \binom{2n}{2} + \ldots + \binom{2n}{2n} \]

Using the Binomial Theorem again, we substitute \(x = 1\):

\[ (1 + 1)^{2n} = 2^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} \]

Thus, we have:

\[ 1 + \binom{2n}{1} + \binom{2n}{2} + \ldots + \binom{2n}{2n} = 2^{2n} \]

**Step 3: Putting it All Together**

Now we can substitute the results back into the original expression:

\[ \frac{(2^{2n})}{(2^{2n})} = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]

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