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The value of C_1 ^2+C_2 ^2....+C_n ^2 (where C_i is the i^{th}

Question

The value of $C_{1}^{2} + C_{2}^{2} .... + C_{n}^{2}$ (where $C_{i}$ is the $i^{t h}$ coefficient of $(1 + x)^{n}$ expansion), is:

\frac{n ^{n}}{n !}

\frac{2 n !}{n ! n !}

\frac{2 n !}{n !}

\frac{n ! \times 2 ^{n}}{2 n !}

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Harper Discussed

The value of

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

(where

C_{i}

is the

i^{t h}

coefficient of

(1 + x)^{n}

expansion), is:

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Text solutionVerified

(1 + x)^{n} =^{n} C_{0} + x (^{n} C_{1}) + x^{2} (^{n} C_{2}) + x^{3} (^{n} C_{3}) + ... + x^{n} (^{n} C_{n})

(x + 1)^{n} = x^{n} (^{n} C_{0}) + x^{n - 1} (^{n} C_{1}) + ... + x^{0} (^{n} C_{n})

Now,

\sum_{r = 0}^{n} (^{n} C_{r})^{2}

= co-efficient of

x^{n}

(1 + x)^{2 n} =^{2 n} C_{n}

Thus,

^{2 n} C_{n} = (^{n} C_{0})^{2} + (^{n} C_{1})^{2} + ... (^{n} C_{n})^{2}

, where

(^{n} C_{i})

is the

i^{t h}

coefficient of

(1 + x)^{n}

expansion.

So,

C_{1}^{2} + C_{2}^{2} + ... + C_{n}^{2} = (^{2 n} C_{n}) = \frac{2 n !}{n ! n !}

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Question Text	The value of $C_{1}^{2} + C_{2}^{2} .... + C_{n}^{2}$ (where $C_{i}$ is the $i^{t h}$ coefficient of $(1 + x)^{n}$ expansion), is:
Answer Type	Text solution:1
Upvotes	150