If n is a positive integer such that `(1+x)^n=^nC_0+^nC_1+^nC_2x^2+…….+^nC_nx^n`, for `epsilonR`. Also `.^nC_r=C_r` On the basis of the above information answer the following questions the value of `.^mC_r.^nC_0+^mC_(r-1).^nC_1+^mC_(r-2).^nC_2+….+^mC_1.^nC_(r-1)+^nC_0^nC_r` where m,n, r are positive interges and `rltm,rltn=`
(A) `.^(mn)C_r`
(B) `.^(m+n)C_r`
(C) 0
(D) 1

If n is a positive integer such that (1+x)^n=^nC_0+^nC_1+^nC_2x^2+…….+^nC_nx^n , for epsilonR . Also .^nC_r=C_r On the basis of the above information answer the following questions The value of the series sum_(r=1)^n r^2.C_r= (A) 1 (B) (-1)^(n/2).(n!)/(n/(2!))^2 (C) (n-1).^(2n)C_n+2(2n) (D) n(n+1)2^(n-2)

If n is a positive integer such that (1+x)^n=^nC_0+^nC_1+^nC_2x^2+…….+^nC_nx^n , for epsilonR . Also .^nC_r=C_r On the basis of the above information answer the following questions The value of the expression a-(a-1)C_1+(a-2)C_2-(a-3)C_3+.......+(1)^n(a-n)C_n= (A) 0 (B) a^n.(-1)^n.^(2n)C_n (C) [2a-n(n+1)].^(2n)C_n (D) none of these

If n is a positive integer then (1+x)^n=^nC_0 x^0+^nC_1 x^1+^nC_2^2+………+^nC_rx^r= sum _(r=0)^n ^nC_rx^r and (1-x)^n= ^nC_0x^0-^nC_1 x^1 +^nC_2-^nC_3x^3+………+(-1)^n ^nC_nx^n=sum_(r=0)^n (-1)^r ^nC_r x^r On the basis of above information answer the following question: If n is a positive integer then 1/((49)^n) - 8/((49)^n)(^(2n)C_1)+8^2/((49)^n)( ^(2n)C_2)- 8^3/((49)^n)(^(2n)c_3)+......+8^(2n)/((49)^n)= (A) -1 (B) 1 (C) (64/49)^n (D) none of these

Statement 1: ^m C_r+^m C_(r-1)(^nC_1)+^mC_(r-2)(^n C_2)+....+^n C_r=0 , if m+n lt r Statement 2: ^n C_r=0 , if n lt r

Show that: .^nC_(n-r)+3.^nC_(n-r+1)+3.^nC_(n-r+2)+^nC_(n-r+3)=^(n+3)C_r .

1/2. ""^nC_0 + ""^nC_1 + 2. ""^nC_2 + 2^2. ""^nC_3 + …….+ 2^(n-1) . ""^nC_n =

Prove that: . n ^C_0+2.^nC_1+…2^n.^nC_n=3^n for every natural number n.

If n is a positive integer and C_r = ""^nC_r then find the value of sum_(x = 1)^n r^2 ((C_r)/(C_(r - 1)) )

Find the sum sum_(r=1)^n r^n (^nC_r)/(^nC_(r-1)) .

If n is a positive integer then ^nC_r+^nC_(r+1)=^(n+1)^C_(r+1) Also coefficient of x^r in the expansion of (1+x)^n=^nC_r. answer the following question: If n is a positive integer then ^nC_n+^(n+1)C_n+^(n+2)C_n+.....+^(n+k)C_n= (A) ^(n+k+1)C_(n+2) (B) ^(n+k+1)C_(n+1) (C) ^(n+k+1)C_k (D) ^(n+k+1)C_(n-2)