If `C_r stands for `^10C_r` show that 2.C_0+2^2/2.C_1+2^3/3.C_2+…+2^11/11.C_10= (3^11-1)/11)`

Prove that 3*""^10C_0+3^2*(""^10C_1)/2+3^3*(""^10C_2)/3+...3^11*(""^10C_10)/11=(4^11-1)/11

2C_(0)+(2^(2))/(2)C_(1)+(2^(3))/(3)C_(2)+..........+(2^(11))/(11)C_(10)=?

11C_(2)r=11C_(r+2)

Find the sum 2C_(0)(23)/(2)quad C_(1)+(2^(3))/(3)C_(2)+(2^(4))/(4)C_(3)+...+(2^(11))/(11)C_(10)

Evaluate : 2^(10)C_(0)+(2^(2).^(10)C_(1))/(2)+(2^(3).^(10)C_(2))/(3)+ . . .+(2^(11).^(10)C_(10))/(11)

Find the sum 2..^(10)C_(0) + (2^(2))/(2).^(10)C_(1) + (2^(3))/(3).^(10)C_(2)+(2^(4))/(4).^(10)C_(3)+"...."+(2^(11))/(11).^(10)C_(10) .

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 (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + …+ C_(n) x^(n) , prove that C_(0)^(2) - C_(1)^(2) + C_(2)^(2) -…+ (-1)^(n) *C_(n)^(2)= 0 or (-1)^(n//2) * (n!)/((n//2)! (n//2)!) , according as n is odd or even Also , evaluate C_(0)^(2) + C_(1)^(2) + C_(2)^(2) - ...+ (-1)^(n) *C_(n)^(2) for n = 10 and n= 11 .

The value of sum_(r=0)^(10)(r)^(20)C_(r) is equal to 20(2^(18)+^(19)C_(10)) b.10(2^(18)+^(19)C_(10)) c.20(2^(18)+^(19)C_(11)) d.10(2^(18)+^(19)C_(11))