Question

Following question contains statements given in two columns which have to be matched The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s Any given statement in Column-I can have correct matching with ONE statement in Column-II

Column-I

Column-II

Answer 1

(A)
$(2n+1)(2n+3)(2n+5).....(4n-1)$
$\frac{(2n!)(2n+1)(2n+2)(2n+3)(2n+4)....(4n-1)\left(4n\right)}{(2n!)(2n+2)(2n+4)(2n+6)....\left(4n\right)}$
$\frac{(4n!)(n!)}{(n!)\left(2n\right)!2^n(n+1)(n+2)...\left(2n\right)}$
$\frac{(4n!)(n!)}{2^n\cdot \left(2n\right)!\left(2n\right)!}$
(B)
$\sum _{r=1}^{n}r\frac{\hat{n}{C}_r}{\hat{n}{C}_{r-1}}=\frac{r.n!}{r!(n-r)!}\frac{(r-1)!(n-r+1)!}{n!}=\sum _{r==1}^n(n-r+1)=n(n+1)-\frac{n(n+1)}{2}=\frac{n(n+1)}{2}$
(C)
$(C_0+C_1)(C_1+C_2)(C_2+C_3).....(C_{n-1}+C_n)=m{C}_1{C}_2...C_{n-1}$
$(\frac{C_0}{C_1}+1)(\frac{C_1}{C_2}+1)(\frac{C_2}{C_3}+1).....(\frac{C_{n-1}}{C_n}+1)=m$
$\frac{n+1}{n}\cdot \frac{n+1}{n-1}\cdot \frac{n+1}{n-2}....\frac{n+1}{1}=m$
$\frac{{(n+1)}^n}{n!}=m$
(D)
$\sum _{i=1}^n\{\sum _{j=1}^{n}i{C}_i{C}_j+\sum _{j=1}^{n}j{C}_i{C}_j\}$
$\sum _{i=1}^{n}i{C}_i(2^n-1)+\sum _{i=1}^n{C}_i\cdot 2^{n-1}$
$=n{.2}^{n-1}(2^n-1)+n\cdot 2^{n-1}(2^n-1)=n\cdot 2^n(2^n-1)$