Prove that `1/(n+1)=(^n C_1)/2-(2(^n C_2))/3+(3(^n C_3))/4-+(-1)^(n+1)(n(^n C_n))/(n+1)` .

Prove that (C_(1))/(1)-(C_(2))/(2)+(C_(3))/(3)-(C_(4))/(4)+...+((-1)^(n-1))/(n)C_(n)=1+(1)/(2)+(1)/(3)+...+(1)/(n)

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

if (1+a)^(n)=.^(n)C_(0)+.^(n)C_(1)a++.^(n)C_(2)a^(2)+ . . .+.^(n)C_(n)a^(n) , then prove that (.^(n)C_(1))/(.^(n)C_(0))+(2(.^(n)C_(2)))/(.^(n)C_(1))+(3(.^(n)C_(3)))/(.^(n)C_(2))+. . . +(n(.^(n)C_(n)))/(.^(n)C_(n-1))= Sum of first n natural numbers.

Prove that (.^(n)C_(0))/(1)+(.^(n)C_(2))/(3)+(.^(n)C_(4))/(5)+(.^(n)C_(6))/(7)+"....."+= (2^(n))/(n+1)

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

Prove that C_(1)^(2)-2*C_(2)^(2)+3*C_(3)^(2)-…-2n*C_(2n)^(2)=(-1)^(n)n*C_(n)

Prove that (^(2n)C_0)^2+(^(2n)C_1)^2+(^(2n)C_2)^2-+(^(2n)C_(2n))^2-(-1)^n^(2n)C_ndot

Find .^(n)C_(1)-(1)/(2).^(n)C_(2)+(1)/(3).^(n)C_(3)- . . . +(-1)^(n-1)(1)/(n).^(n)C_(n)

C_(0)-(C_(1))/(2)+(C_(2))/(3)-......+(-1)^(n)(C_(n))/(n+1)=(1)/(n+1)