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Question Number 23856 by Tinkutara last updated on 08/Nov/17

The value of (C_{0} + C_{1}) (C_{1} + C_{2}) \dots .

(C_{n - 1} + C_{n}) is

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

(2) \frac{{(n - 1)}^{n}}{n!} \cdot C_{1} C_{2} \dots . . C_{n}

(3) \frac{{(n)}^{n}}{(n + 1)!} \cdot C_{1} C_{2} \dots . . C_{n}

(4) \frac{{(n)}^{n}}{n!} \cdot C_{1} C_{2} \dots . . C_{n}

Commented by prakash jain last updated on 08/Nov/17

C_{k} + C_{k + 1} = \frac{n!}{k! (n - k)!} + \frac{n!}{(k + 1)! (n - k - 1)!}

= \frac{n!}{k! (n - k - 1)!} [\frac{n - k + k + 1}{(k + 1) (n - k)}]

= \frac{n + 1!}{(k + 1)! (n - k)!} = \frac{(n + 1) (n!)}{(k + 1) [k! (n - k)!]}

= \frac{(n + 1)}{(k + 1)}^{n} C_{k}

Commented by prakash jain last updated on 08/Nov/17

\underset{k = 0}{\prod^{n - 1}} (C_{k} + C_{k + 1})

= \underset{k = 0}{\prod^{n - 1}} \frac{n + 1}{k + 1} C_{k}

= \frac{{(n + 1)}^{n}}{n!} C_{0} C_{1} \dots C_{n - 1}

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

Commented by prakash jain last updated on 08/Nov/17

Option (1) is correct .

Commented by Tinkutara last updated on 09/Nov/17

Thank you very much Sir!