If (1 + x)^n = sumlimits_{r = 0}^n {{C_r}{x^r | vedclass

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(C) Given the expression $P = \left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{

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$\frac{{{{(n + 1)}^n}}}{{n!}}$

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(C) Given the expression $P = \left( {1 + \frac{{{C_1}}}{{{C_0}}}} ight)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} ight)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} ight)$.We know that $\frac{{{C_r}}}{{{C_{r - 1}}}} = \frac{{n - r + 1}}{r}$.Substituting this into each term: $1 + \frac{{{C_r}}}{{{C_{r - 1}}}} = 1 + \frac{{n - r + 1}}{r} = \frac{{r + n - r + 1}}{r} = \frac{{n + 1}}{r}$.Thus,the product becomes:$P = \left( \frac{n + 1}{1} ight) \left( \frac{n + 1}{2} ight) \left( \frac{n + 1}{3} ight) .... \left( \frac{n + 1}{n} ight)$.$P = \frac{{{{(n + 1)}^n}}}{{1 	imes 2 	imes 3 	imes .... 	imes n}} = \frac{{{{(n + 1)}^n}}}{{n!}}$.

If $(1 + x)^n = \sum\limits_{r = 0}^n {{C_r}{x^r}} $,then $\left( {1 + \frac{{{C_1}}}{{{C_0}}}} \right)\left( {1 + \frac{{{C_2}}}{{{C_1}}}} \right)....\left( {1 + \frac{{{C_n}}}{{{C_{n - 1}}}}} \right) = $

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