यदि (1+x)^(20) के प्रसार में C_0,C_1,C_2....,C_(20) द्विपद गुणांक हैं, तो c_1/c_0+2c_2/c_1+3c_3/...

C(20,0)+C(20,1)+C(20,2)+...+C(20,10)

1.^(20)C_(1)-2.^(20)C_(2)+3.^(20)C_(3)-...-20.^(20)C_(20)

If (1+x)^(n)=C_(0)+C_(1)x+…..+C_(n)x^(n) , then (C_(1))/(C_(0))+(2C_(2))/(C_(1))+(3C_(3))/(C_(2))+....+(nC_(n))/(C_(n-1)) is :

If (1+x)^n=C_0+C_1x+C_2x^2+C_3x^3+...+C_nx^n then prove that 2.C_0+2^2C_1/2+2^3C_2/3+2^4C_3/4+...+2^(n+1)C_n/(n+1)=(3^(n+1)-1)/(n+1)

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

If (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + ... + C_(n) x^(n) , then value of C_(0)^(2) + 2C_(1)^(2) + 3C_(2)^(2) + ... + (n + 1) C^(2)n is

If (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3)x^(3) + …+ C_(n) x^(n) , then C_(0) - (C_(0) - C_(1)) + (C_(0) + C_(1) + C_(2))- (C_(0) + C_(1) + C_(2)+ C_(3)) + ...+ (-1)^(n-1) (C_0) + C_(1) + C_(2) + ...+ C_(n-1)) , when n is even integer is

If C_(0)+C_(1)+C_(2)+......+C_(n)=128 then C_(0)-(C_(1))/(2)+(C_(2))/(3)-(C_(3))/(4)+.....=

If C_(0)+C_(1)+C_(2)+......+C_(n)=128 then C_(0)-(C_(1))/(2)+(C_(2))/(3)-(C_(3))/(4)+......=