Statement 1: `^m C_r+^m C_(r-1)^n C_1+^mC_(r-2)^n C_2++^n C_r=0,ifm+n

Prove that (r+1)^n C_r-r^n C_r+(r-1)^n C_2-^n C_3++(-1)^r^n C_r = (-1)^r^(n-2)C_rdot

The value of determinant |[ ^n C_(r-1), ^n C_r, (r+1)^(n+2)C_(r+1)],[ ^n C_r, ^n C_(r+1),(r+2)^(n+2)C_(r+2)],[ ^n C_(r+1), ^n C_(r+2), (r+3)^(n+2)C_(r+3)]| is n^2+n-2 b. 0 c. ^n+3C_(r+3) d. ^n C_(r-1)+^n C_r+^n C_(r+1)

Show that |^x C_r^x C_(r+1)^x C_(r+2)^y C_r^y C_(r+1)^y C_(r+2)^z C_r^z C_(r+1)^z C_(r+1)|=|^x C_r^(x+1)C_(r+1)^(x+2)C_(r+2)^y C_r^(y+1)C_(r+1)^(y+2)C_(r+2)^z C_r^(z+1)C_(r+1)^(z+2)C_(r+1)| .

Using binomial theorem (without using the formula for ^n C_r ) , prove that "^n C_4+^m C_2-^m C_1^n C_2 = ^m C_4-^(m+n)C_1^m C_3+^(m+n)C_2^m C_2-^(m+n)C_3^m C_1+^ (m+n)C_4dot

Prove that ^n C_0^n C_0-^(n+1)C_1^n C_1+^(n+2)C_2^n C_2-=(-1)^ndot

If (1+2x+x^2)^n=sum_(r=0)^(2n)a_r x^r ,then a= (^n C_2)^2 b. ^n C_rdot^n C_(r+1) c. ^2n C_r d. ^2n C_(r+1)

Let m, in N and C_(r) = ""^(n)C_(r) , for 0 le r len Statement-1: (1)/(m!)C_(0) + (n)/((m +1)!) C_(1) + (n(n-1))/((m +2)!) C_(2) +… + (n(n-1)(n-2)….2.1)/((m+n)!) C_(n) = ((m + n + 1 )(m+n +2)…(m +2n))/((m +n)!) Statement-2: For r le 0 ""^(m)C_(r)""^(n)C_(0)+""^(m)C_(r-1)""^(n)C_(1) + ""^(m)C_(r-2) ""^(n)C_(2) +...+ ""^(m)C_(0)""^(n)C_(r) = ""^(m+n)C_(r) .

The roots of the equation |^x C_r^(n-1)C_r^(n-1)C_(r-1)^(x+1)C_r^n C_r^n C_(r-1)^(x+2)C_r^(n+1)C_r^(n+1)C_(r-1)|=0 are x=n b. x=n+1 c. x=n-1 d. x=n-2

Find the sum of sum_(r=1)^n(r^n C_r)/(n C_(r-1) .

Prove that "^n C_0^(2n)C_n-^n C_1^(2n-1)C_n+^n C_2xx^(2n-2)C_n++(-1)^n^n C_n^n C_n=1.