ISI2014-DCG-1

Question

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

• A. $\left( \frac{n+1}{n+2} \right) ^n$
• B. $ \frac{n^n}{n!} $
• C. $\left( \frac{n}{n+1} \right) ^n$
• D. $ \frac{(n+1)^n}{n!} $

Answer 1

(D) $\frac{(n+1)^n}{n!}$

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

Then,$\frac{C_p}{C_{p+1}}=\frac{p+1}{n-p}$

Then,$1+\frac{C_p}{C_{p+1}}=\frac{n+1}{n-p}$

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 $\frac{(n+1)^n}{n!}$

Comments

Put n=1 then(1+1)=2

Check in answer Option D satisfied

Answer 2

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.