First slide

Binomial theorem for positive integral Index

Question

Let $P_{n}$ denote the product of all the coefficients in the expansion of expansion of ${(1 + x)}^{n} .$ If $(20)! P_{n + 1} = 21^{20} P_{n},$ then $n$ is equal to

Moderate

A

21
B

20
C

19
D

18

Solution

Suppose
$(1 + x)^{n + 1} = \sum_{k = 0}^{n + 1} B_{k} x^{k},$ where $B_{k} =^{n + 1} C_{k}$
and $(1 + x)^{n} = \sum_{k = 0}^{n} C_{k} x^{k},$ where $C_{k} =^{n} C_{k},$
We have
$\frac{P_{n + 1}}{P_{n}} = {\bar{B}}_{0} (\frac{B_{1}}{A_{0}}) (\frac{B_{2}}{A_{1}}) \dots (\frac{B_{n + 1}}{A_{n}})$
Now, $\frac{B_{r + 1}}{A_{r}} = \frac{(n + 1)!}{(r + 1)! (n - r)!} \cdot \frac{r! (n - r)!}{n!}$
$= \frac{n + 1}{r + 1} (0 \leq r \leq n)$
Thus, $\frac{P_{n + 1}}{P_{n}} = \frac{(n + 1)^{n}}{r!}$
$\Rightarrow \frac{21^{20}}{20!} = \frac{(n + 1)^{n}}{r!} \Rightarrow n = 20$

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