If (frac{1}{^{15}C_{0}}+frac{1}{^{15}C_{1}})( | vedclass

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(B) We know that $\frac{1}{^{n}C_{r}} + \frac{1}{^{n}C_{r+1}} = \frac{r! (n-r)!}{n!} + \frac{(r+1)! (n-r-1)!}{n!} = \frac{(n-r-1)! r!}{n!} [n-r + r+1]

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) We know that $\frac{1}{^{n}C_{r}} + \frac{1}{^{n}C_{r+1}} = \frac{r! (n-r)!}{n!} + \frac{(r+1)! (n-r-1)!}{n!} = \frac{(n-r-1)! r!}{n!} [n-r + r+1] = \frac{(n+1) (n-r-1)! r!}{n!} = \frac{n+1}{n-r} \cdot \frac{1}{^{n}C_{r}}$.Using the property $\frac{1}{^{n}C_{r}} + \frac{1}{^{n}C_{r+1}} = \frac{n+1}{n+1} \cdot \frac{^{n+1}C_{r+1}}{^{n}C_{r} \cdot ^{n}C_{r+1}} = \frac{n+1}{n+1} \cdot \frac{1}{^{n}C_{r}} \cdot \frac{n+1}{r+1} = \frac{16}{15} \cdot \frac{1}{^{14}C_{r}}$.Thus, the product is $\prod_{r=0}^{12} (\frac{1}{^{15}C_{r}} + \frac{1}{^{15}C_{r+1}}) = \prod_{r=0}^{12} (\frac{16}{15} \cdot \frac{1}{^{14}C_{r}}) = \frac{(16/15)^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$.Comparing with the given expression, $a = \frac{16}{15}$.Therefore, $30a = 30 \cdot \frac{16}{15} = 32$.

If $(\frac{1}{^{15}C_{0}}+\frac{1}{^{15}C_{1}})(\frac{1}{^{15}C_{1}}+\frac{1}{^{15}C_{2}})...(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}) = \frac{a^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$, then $30a$ is equal to:

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