n का मान निकालिए यदि
` 2n ! n ! = (n+1) (n-1)! (2n-1) ! `

Find the value of n if: 2n! n! =(n+1)(n-1)(!(2n-1)!

Find n if : ""^(2n+1) P_(n-1) : ""^(2n-1) P_n = 3: 5

((n+2)!+(n+1)(n-1)!)/((n+1)(n-1)!)=

For each n in N, n(n+1) (2n+1) is divisible by

For each n in N, n(n+1) (2n+1) is divisible by

If n in N, n > 1 , then value of E= a - ""^(n)C_(1) (a-1) + ""^(n)C_(2) (a -2)+ ... + (- 1)^(n) (a-n) (""^(n)C_(n)) is

Sum of the series S_(n) =(n) (n) + (n-1) (n+1) + (n-2) (n+2) + …+ 1(2n-1) is

The coefficient of 1//x in the expansion of (1+x)^n(1+1//x)^n is (n !)/((n-1)!(n+1)!) b. ((2n)!)/((n-1)!(n+1)!) c. ((2n)!)/((2n-1)!(2n+1)!) d. none of these