If `sum_(k=1)^m(k^2+1)k! =1999(2000 !)` , then m is

sum_(k=1)^m(k^2+1)k=2009xx2010!, then m= (A) 2009 (B) 2010 (C) 2011 (D) none of these

sum_(k=1)^m(k^2+1)k=2009xx2010!, then m= (A) 2009 (B) 2010 (C) 2011 (D) none of these

if underset(k=1)oversetnsum(k^2+1)k! =1999xx2000! then prove that n=1999

sum_(k=0)^(5)(-1)^(k)2k

sum_(k=1)^(2n+1) (-1)^(k-1) k^2 =

If z_(k)=e^(i theta_k) for k= 1, 2, 3, 4 where i^(2)= -1 , and if |sum_(k=1)^(4) (1)/(z_k)|=1 , then |sum_(k=1)^(4)| is equal to

sum_(k =1)^(n) k(1 + 1/n)^(k -1) =

sum_(k=1)^(2n+1)(-1)^(k-1)k^(2)=

Suppose m and n are positive integers and let S=sum_(k=0)^(n)(-1)^(k)(1)/(k+m+1)(nC_(k)) and T=sum_(k=0)^(m)(-1)^(k)1(k+n+1)(mC_(k)) then S-T is equal to