If `\ ^m C_1=\ \ ^n C_2` then `2m=n` b. `2m=n(n+1)` c. `2m=(n-1)` d. `2n=m(m-1)`

If \ ^m C_1=\ \ ^n C_2 then which is correct a. 2m=n b. 2m=n(n+1) c. 2m=(n-1) d. 2n=m(m-1)

If ^(m)C_(1)=^(n)C_(2) then 2m=nb2m=n(n+1) c.2m=(n-1)d2n=m(m-1)

If .^(m)C_(1)=.^(n)C_(2) prove that m=(1)/(2)n(n-1) .

If S_n denotes the sum of first n terms of an A.P. such that (S_m)/(S_n)=(m^2)/(n^2),\ t h e n(a_m)/(a_n)= a. (2m+1)/(2n+1) b. (2m-1)/(2n-1) c. (m-1)/(n-1) d. (m+1)/(n+1)

If S_n denotes the sum of first n terms of an A.P. such that (S_m)/(S_n)=(m^2)/(n^2), t h e n(a_m)/(a_n)= a.(2m+1)/(2n+1) b. (2m-1)/(2n-1) c. (m-1)/(n-1) d. (m+1)/(n+1)

Let m ,n are integers with 0

The value of ^n C_1+^(n+1)C_2+^(n+2)C_3++^(n+m-1)C_m is equal to (a) ^m+n C_(n-1) (b) ^m+n C_(n-1) (c) ^mC_(1)+^(m+1)C_2+^(m+2)C_3++^(m+n-1) (d) ^m+1C_(m-1)

Let T_r a n dS_r be the rth term and sum up to rth term of a series, respectively. If for an odd number n ,S_n=na n dT_n=(T_n-1)/(n^2),t h e nT_m ( m being even)is 2/(1+m^2) b. (2m^2)/(1+m^2) c. ((m+1)^2)/(2+(m+1)^2) d. (2(m+1)^2)/(1+(m+1)^2)

Let T_r a n dS_r be the rth term and sum up to rth term of a series, respectively. If for an odd number n ,S_n=na n dT_n=(T_n-1)/(n^2),t h e nT_m ( m being even)is 2/(1+m^2) b. (2m^2)/(1+m^2) c. ((m+1)^2)/(2+(m+1)^2) d. (2(m+1)^2)/(1+(m+1)^2)

The value of .^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"….."+.^(n+m-1)C_(m) is equal to a. .^(m+n)C_(n) - 1 b. .^(m+n)C_(n-1) c. .^(m)C_(1) + ^(m+1)C_(2) + ^(m+2)C_(3) + "…." + ^(m+n-1)C_(n) d. .^(m+n)C_(m) - 1