The value of ${({}^{n}{C_{1}})}^{2} + {({}^{n}{C_{2}})}^{2} + {({}^{n}{C_{3}})}^{2} . . . . . . . + {({}^{n}{C_{n}})}^{2}$ is

${({}^{2 n}{C_{n}})}^{2}$

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${}^{2 n}{C_{n}}$

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${}^{2 n}{C_{n} + 1}$

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${}^{2 n}{C_{n} - 1}$

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Solution

The correct option is D
${}^{2 n}{C_{n} - 1}$

Explanation for the correct answer:
Step 1: Apply Binomial Theorem
From the Binomial theorem, we can write,
${(1 + x)}^{n} = {}^{n}{C_{0} + {}^{n}{C_{1} x} + {}^{n}{C_{2} x^{2}} + . . . . . + {}^{n}{C_{n}}} x^{n}$ $. . . (i)$
${(x + 1)}^{n} = {}^{n}{C_{0} x^{n} + {}^{n}{C_{1} x^{n - 1}} + {}^{n}{C_{2} x^{n - 2}} + . . . . . + {}^{n}{C_{n}}}$ $. . . (i i)$
Multiplying $(i)$ and $(i i)$ we get,
${(1 + x)}^{n} {(x + 1)}^{n} = ({}^{n}{C_{0} + {}^{n}{C_{1} x} + {}^{n}{C_{2} x^{2}} + . . . . . + {}^{n}{C_{n}}} x^{n}) \times ({}^{n}{C_{0} x^{n} + {}^{n}{C_{1} x^{n - 1}} + {}^{n}{C_{2} x^{n - 2}} + . . . . . + {}^{n}{C_{n}}})$
$\Rightarrow$ ${(1 + x)}^{2 n} = ({}^{n}{C_{0} + {}^{n}{C_{1} x} + {}^{n}{C_{2} x^{2}} + . . . . . + {}^{n}{C_{n}}} x^{n}) \times ({}^{n}{C_{0} x^{n} + {}^{n}{C_{1} x^{n - 1}} + {}^{n}{C_{2} x^{n - 2}} + . . . . . + {}^{n}{C_{n}}})$
Comparing the coefficients of $x^{n}$ on both sides of the equation
Coefficient of $x^{n}$ on RHS $= {({}^{n}{C_{0}})}^{2} + {({}^{n}{C_{1}})}^{2} + {({}^{n}{C_{2}})}^{2} + . . . . . . . + {({}^{n}{C_{n}})}^{2}$
Step 2: Apply formula for coefficient of a term in binomial expansion
Coefficient of $x^{n}$ on LHS $= {}^{2 n}{C_{n}}$
$\Rightarrow$ ${}^{2 n}{C_{n} =} {({}^{n}{C_{0}})}^{2} + {({}^{n}{C_{1}})}^{2} + {({}^{n}{C_{2}})}^{2} + . . . . . . . + {({}^{n}{C_{n}})}^{2}$
$\Rightarrow$ ${}^{2 n}{C_{n} =} {(\frac{n!}{0! n!})}^{2} + {({}^{n}{C_{1}})}^{2} + {({}^{n}{C_{2}})}^{2} + . . . . . . . + {({}^{n}{C_{n}})}^{2}$
$\Rightarrow$ ${}^{2 n}{C_{n} =} 1 + {({}^{n}{C_{1}})}^{2} + {({}^{n}{C_{2}})}^{2} + . . . . . . . + {({}^{n}{C_{n}})}^{2}$
$\Rightarrow$ ${({}^{n}{C_{1}})}^{2} + {({}^{n}{C_{2}})}^{2} + . . . . . . . + {({}^{n}{C_{n}})}^{2} = {}^{2 n}{C_{n} - 1}$
Hence the value of ${({}^{n}{C_{1}})}^{2} + {({}^{n}{C_{2}})}^{2} + {({}^{n}{C_{3}})}^{2} . . . . . . . + {({}^{n}{C_{n}})}^{2}$ is ${}^{2 n}{C_{n} - 1}$ .
Hence, option $(D)$ is the correct answer.

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