If `.^(20)C_(1)+(2^(2)) .^(20)C_(2) + (3^(2)).^(20)C_(3) +......+(20^(2)).^(20)C_(20)=A(2^(beta))` , then the ordered pair `(A, beta)` is equal toA. `(420 , 19)`B. `(420 , 18)`C. `(380, 18)`D. ` (380, 19)`

If `.^(20)C_(1)+(2^(2)) .^(20)C_(2) + (3^(2)).^(20)C_(3) +......+(20^(

1.	If `.^(20)C_(1)+(2^(2)) .^(20)C_(2) + (3^(2)).^(20)C_(3) +......+(20^(2)).^(20)C_(20)=A(2^(beta))` , then the ordered pair `(A, beta)` is equal toA. `(420 , 19)`B. `(420 , 18)`C. `(380, 18)`D. ` (380, 19)`
Answer» Correct Answer - B We Know, `(1+x)^(n) = .^(n)C_(0) + .^(n)C_(1)x + .^(n)C_(2) x^(2) +... +.^(n)C_(n) x^(n) ` On differentiating both sides w. r. t. X, we get `n(1+x)^(n-1)=.^(n)C_(1) + 2 .^(n)C_(2) x + ... + n .^(n)C_(n) x^(n-1)` On multiplying both sides by x, we get `n x(1+x)^(n-1)=.^(n)C_(1)x+2^(n) C_(2)x^(2)+...+n .^(n)C_(n) x^(n) ` Again on differentiating both sides w.r.t. x, we get `n[(1+x)^(n-1) +(n-1) x (1+x)^(n-2)]=.^(n)C_(1) + 2^(2) .^(n)C_(2)x+...+ n^(2) .^(n)C_(n) x^(n-1) ` Now putting x = 1 in both sides, we get `.^(n)C_(1) +(2^(2)).^(n)C_(2)+(3^(2)) .^(n)C_(3) + ...+(n^(2)).^(n)C_(n) = n(2^(n-1)+(n-1) 2^(n-2))` For n = 20, we get `.^(20)C_(1) +(2^(2)) .^(20)C_(2)+(3^(2)) .^(20)C_(3) + ...+ (20)^(2) .^(20)C_(20) ` ` = 20 (2^(19)+(19) 2^(18))` ` = 20 (2+19) 2^(18) = 420 (2^(18))` ` = A(2^(B))` (given) On comparing, we get (A,B) = (420, 18)