If ${(1 + x)}^{n} = \sum_{r = 0}^{n}{a_{r}x^{r}}$ and $b_{r} = 1 + \frac{a_{r}}{a_{r} - 1}$ and $\prod_{r = 1}^{n}{b_{r} = \frac{{(101)}^{100}}{100!}}$, then $n$ is

Question

Dr. Abubacker · Accepted Answer

(b)
$\because\ {(1 + x)}^{n} = \sum_{r = 0}^{n}{\ ^{n}C_{r}x^{r} =}\sum_{r = 0}^{n}{a_{r}x^{r}}$ (given)
$	herefore\ a_{r} = \ ^{n}C_{r}$
Also, $b_{r} = 1 + \frac{a_{r}}{a_{r} - 1} = 1 + \frac{\ ^{n}C_{r}}{\ ^{n}C_{r - 1}} = \frac{\ ^{n + 1}C_{r}}{\ ^{n}C_{r - 1}}$
$b_{r} = \left( \frac{n + 1}{r} ight)$
$\because\ \prod_{r = 1}^{n}{b_{r} = \prod_{r = 1}^{n}{\left( \frac{n + 1}{r} ight) = \frac{(n + 1)^{n}}{n!}}}$
$= \frac{{(101)}^{100}}{100!}$ (given)
$	herefore\ n = 100$

Q. If ${(1 + x)}^{n} = \sum_{r = 0}^{n}{a_{r}x^{r}}$ and $b_{r} = 1 + \frac{a_{r}}{a_{r} - 1}$ and $\prod_{r = 1}^{n}{b_{r} = \frac{{(101)}^{100}}{100!}}$, then $n$ is

Explanation

Reviewed by Dr. Abubacker

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