Question

If ${(1+x)}^n=\sum _{r=0}^n{}^n{C}_r$ then solve

$\left(\frac{C_0+C_1}{C_0}\right)\left(\frac{C_1+C_2}{C_1}\right)\left(\frac{C_2+C_3}{C_2}\right).........\left(\frac{C_{n-1}+C_n}{C_{n-1}}\right)$

Answer 1

If ${(1+x)}^n=\sum _{r=0}^n{}^n{C}_r$ then find

$\left(\frac{C_0+C_1}{C_0}\right)\left(\frac{C_1+C_2}{C_1}\right)\left(\frac{C_3+C_2}{C_2}\right)...............\left(\frac{C_{n-1}+C_n}{C_n}\right)$

We can be written this

$(1+\frac{C_1}{C_0})\left(\frac{C_1+C_2}{C_1}\right)(1+\frac{C_2}{C_1})(1+\frac{C_3}{C_2}).............(1+\frac{C_n}{C_{n-1}})$

We know that

$\frac{C_r}{C_{r-1}}=\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}$

$\frac{C_r}{C_{r-1}}=\frac{n!}{r!(n-r)!}\times \frac{(r-1)!(n-r+1)!}{n!}$

$\frac{C_r}{C_{r-1}}=\frac{n!}{r(r-1)!(n-r)!}\times \frac{(r-1)!(n-r+1)(n-r)!}{n!}$

$\frac{C_r}{C_{r-1}}=\frac{n-r+1}{r}$

Then,

$(1+\frac{C_r}{C_{r-1}})=1+\frac{n-r+1}{r}$

$=\frac{r+n-r+1}{r}$

$=\frac{n+1}{r}$

Therefore,

$(1+\frac{C_1}{C_0})\left(\frac{C_1+C_2}{C_1}\right)(1+\frac{C_2}{C_1})(1+\frac{C_3}{C_2}).............(1+\frac{C_n}{C_{n-1}})$

$=\frac{n+1}{1}.\frac{n+1}{2}.\frac{n+1}{3}........\frac{n+1}{n}$

$=\frac{{(n+1)}^n}{n}$

Hence, it is required solution.