Find the value of `1/(81^n)-(10)/(81^n)^(2n)C_1+(10^2)/(81^n)^(2n)C_2-(10^3)/(81^n)^(2n)C_3++(10^(2n))/(81^n)`

If the value of "^(n)C_(0)+2*^(n)C_(1)+3*^(n)C_(2)+...+(n+1)*^(n)C_(n)=576 , then n is

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

The value of ""^(n)P_(1)+(""^(n)P_(2))/(2!)+(""^(n)P_(3))/(3!)+......+(""^(n)P_(n))/(n!)

In a n- sided regular polygon, the probability that the two diagonal chosen at random will intersect inside the polygon is (2^n C_2)/(^(^(n C_(2-n)))C_2) b. (^(n(n-1))C_2)/(^(^(n C_(2-n)))C_2) c. (^n C_4)/(^(^(n C_(2-n)))C_2) d. none of these

The value of .^(n)C_(0) xx .^(2n)C_(r) - .^(n)C_(1)xx.^(2n-2)C_(r)+.^(n)C_(2)xx.^(2n-4)C_(r)+"…." is equal to

Two players P_1a n dP_2 play a series of 2n games. Each game can result in either a win or a loss for P_1dot the total number of ways in which P_1 can win the series of these games is equal to a. 1/2(2^(2n)-^ "^(2n)C_n) b. 1/2(2^(2n)-2xx^"^(2n)C_n) c. 1/2(2^n-^"^(2n)C_n) d. 1/2(2^n-2xx^"^(2n)C_n)

Find the sum 1.^(n)C_(0) + 3 .^(n)C_(1) + 5.^(n)C_(2) + "….." + (2n+1).^(n)C_(n) .

The value of ("^n C_0)/n + ("^nC_1)/(n+1) + ("^nC_2)/(n+2) +....+ ("^nC_ n)/(2n) is equal to

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