`[(^n C_0+^n C_3+)1//2(^n C_1+^n C_2+^n C_4+^n C_5]^2+3//4(^n C_1-^n C_2+^n C_4-^n C_5+)^2=` `3` b. `4` c. `2` d. `1`

The value of |1 1 1\ ^n C_1\ ^(n+2)C_1\ ^(n+4)C_1\ ^n C_2\ ^(n+2)C_2\ ^(n+4)C_2| is (a) 2 (b) 4 (c) 8 (d) n^2

If n=5 ,then ("^n C_0)^(2) + ("^n C_1)^(2) + ("^n C_2)^(2) +......+ ("^n C_5)^(2) is equal to

2C_0 + 2^2 (C_1)/(2) + 2^3 (C_2)/(3) + ………. + 2^(n+1) (C_n)/(n+1) = (3^(n+1) - 1)/(n+1)

Prove : C_0 + 1/3C_2 + 1/5 C_4 + 1/7 C_6 + ………… = (2^n)/(n+1)

If ^(n+1)C_(3)=2.quad ^(n)C_(2) then =3 b.4c.6d.5

(1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3) x^(3) + … + C_(n) x^(n) , prove that C_(0) - 2C_(1) + 3C_(2) - 4C_(3) + … + (-1)^(n) (n+1) C_(n) = 0

If (1+x)^n=C_0+C_1x+C_2x^2+…+C_nx^n show that C_1-2C_2+3C_3-4C_4+…+(-1)^(n-1) n.C_n=0 where C_r=^nC_r .

If ^(n+1)C_3=2.^nC_2, then n= (A) 3 (B) 4 (C) 5 (D) 6