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If $P_n$ denotes the product of binomial coeff. in the expansion of $(1+x)^n$, then $\large\frac{P_{n+1}}{P_n}$=?

$\begin{array}{1 1}=\large\frac{(n+1)^n}{n!} \\ =\large\frac{(n+1)^n}{(n+1)!} \\=\large\frac{(n+1)^{n+1}}{n!} \\ =\large\frac{n^n}{n!} \end{array} $

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Toolbox:
  • $^nC_r=\large\frac{n}{r}$$\times ^{n-1}C_{r-1}$
  • $^{n+1}C_0=1$
$P_n=^nC_0.^nC_1.^nC_2..............^nC_n$
$P_{n+1}=^{n+1}C_0.^{n+1}C_1.^{n+1}C_2.......^{n+1}C_{n+1}$
$\therefore\:\large\frac{P_{n+1}}{P_n}=\frac{^{n+1}C_0.^{n+1}C_1.^{n+1}C_2.......^{n+1}C_{n+1}}{^nC_0.^nC_1.^nC_2..............^nC_n}$
$=\bigg(\large\frac{^{n+1}C_1}{^nC_0}\bigg).\bigg(\large\frac{^{n+1}C_2}{^nC_1}\bigg).............\bigg(\large\frac{^{n+1}C_{n+1}}{^nC_n}\bigg)$
$=\bigg(\large\frac{\frac{n+1}{1}.^nC_0}{^nC_0}\bigg)$.$\bigg(\large\frac{\frac{n+1}{2}.^nC_1}{^nC_1}\bigg)$................$\bigg(\large\frac{\frac{n+1}{n+1}.^nC_n}{^nC_n}\bigg)$
$=\large\frac{n+1}{1}.\frac{n+1}{2}........\frac{n+1}{n+1}$
$=\large\frac{(n+1)^n}{n!}$
answered Oct 14, 2013 by rvidyagovindarajan_1
 

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