Solveeit Logo
Login

Question

Question: If \((1 + x)^{n} = \sum_{r = 0}^{n}{C_{r}x^{r}}\), then \(\left( 1 + \frac{C_{1}}{C_{0}} \right)\lef...

If (1+x)n=∑r=0nCrxr(1 + x)^{n} = \sum_{r = 0}^{n}{C_{r}x^{r}}, then (1+C1C0)(1+C2C1)......(1+CnCn−1)=\left( 1 + \frac{C_{1}}{C_{0}} \right)\left( 1 + \frac{C_{2}}{C_{1}} \right)......\left( 1 + \frac{C_{n}}{C_{n - 1}} \right) =

A

nn−1(n−1)!\frac{n^{n - 1}}{(n - 1)!}

B

(n+1)n−1(n−1)!\frac{(n + 1)^{n - 1}}{(n - 1)!}

C

(n+1)nn!\frac{(n + 1)^{n}}{n!}

D

(n+1)n+1n!\frac{(n + 1)^{n + 1}}{n!}

Answer

(n+1)nn!\frac{(n + 1)^{n}}{n!}

Explanation

Solution

We have (1+C1C0)(1+C2C1)........(1+CnCn−1)\left( 1 + \frac{C_{1}}{C_{0}} \right)\left( 1 + \frac{C_{2}}{C_{1}} \right)........\left( 1 + \frac{C_{n}}{C_{n - 1}} \right)

= (1+n1)(1+n(n−1)/2!n)......(1+1n)\left( 1 + \frac{n}{1} \right)\left( 1 + \frac{n(n - 1)/2!}{n} \right)......\left( 1 + \frac{1}{n} \right)

= (1+n1)(1+n2)(1+n3).......(1+nn)=(n+1)nn!\left( \frac{1 + n}{1} \right)\left( \frac{1 + n}{2} \right)\left( \frac{1 + n}{3} \right).......\left( \frac{1 + n}{n} \right) = \frac{(n + 1)^{n}}{n!}

Trick : Put

n=1,2,3.......,S1=1+1C11C0=2,S2=(1+2C12C1)(1+2C22C1)=92n = 1,2,3.......,S_{1} = 1 + \frac{1C_{1}}{1C_{0}} = 2,S_{2} = \left( 1 + \frac{2C_{1}}{2C_{1}} \right)\left( 1 + \frac{2C_{2}}{2C_{1}} \right) = \frac{9}{2}Which is given by option (3) n=1n = 1, (1+1)11!=2\frac{(1 + 1)^{1}}{1!} = 2;

For n=2n = 2, (2+1)22!=92\frac{(2 + 1)^{2}}{2!} = \frac{9}{2}