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

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

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

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

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

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

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

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

Find sum_(k=1)^(n)(1)/(k(k+1)) .