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+2 votes

Let $(1+x)^n = C_0+C_1x+C_2x^2+ \dots + C_nx^n$, $n$ being a positive integer. The value of

$$\left( 1+\dfrac{C_0}{C_1} \right) \left( 1+\dfrac{C_1}{C_2} \right) \cdots \left( 1+\dfrac{C_{n-1}}{C_n} \right)$$ is

  1. $\left( \frac{n+1}{n+2} \right) ^n$
  2. $ \frac{n^n}{n!} $
  3. $\left( \frac{n}{n+1} \right) ^n$
  4. $ \frac{(n+1)^n}{n!} $
in Combinatory by Veteran (431k points)
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2 Answers

+7 votes

(D) $\frac{(n+1)^n}{n!}$

Here $C_p=$$\binom{n}{p}$



For p=0 $=>$ $1+\frac{C_0}{C_{1}}=\frac{n+1}{n}$

For p=1 $=>$ $1+\frac{C_1}{C_{2}}=\frac{n+1}{n-1}$

and so on..

Thus if we multiply all the terms we get $\frac{(n+1)^n}{n!}$

by Active (1.7k points)
Put n=1 then(1+1)=2

Check in answer Option D satisfied
0 votes
If we take the value of N==3, then from the condition we get the value of expression as 32/3.


Now checking the options when we put N==3 we must get the same value and this is satisfied only by option D put n==3 we get 4^3/Fact(3) which is equal to 32/3 hence the correct option is Option D.
by (289 points)

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